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# Cubic Tap Code
This works similarly to [Tap Code](https://en.wikipedia.org/wiki/Tap_code) except instead of being mapped onto a 5x5 square, letters are mapped onto a 3x3x3 cube, left to right, top to bottom, front to back with space being the 27th "letter". Letters are represented by a series of taps (represented as dots `.`) and pauses (represented by spaces ` `), for example `A` is represented as `. . .` (first column, first row, first layer) and ` ` is represented as `... ... ...` (third column, third row, third layer).
For reference the three layers of the cube are as follows (underscore represents space):
```
1 1 2 3
1 A B C
2 D E F
3 G H I
2 1 2 3
1 J K L
2 M N O
3 P Q R
3 1 2 3
1 S T U
2 V W X
3 Y Z _
```
### Your task (should you choose to accept it)
Create two functions `encode()` and `decode()`, to encode and decode strings to and from cubic tap code.
### Input
`encode()` takes a string of uppercase letters and spaces and outputs a string of dots and spaces.
`decode()` takes a string of dots and spaces and outputs a string of uppercase letters and spaces.
All inputs will be valid.
### Examples
```
encode("N") => ".. .. .."
encode("TEST") => ".. . ... .. .. . . . ... .. . ..."
encode("HELLO WORLD") => ".. ... . .. .. . ... . .. ... . .. ... .. .. ... ... ... .. .. ... ... .. .. ... ... .. ... . .. . .. ."
decode(".. .. ..") => "N"
decode(".. . ... .. .. . . . ... .. . ...") => "TEST"
decode(".. ... . .. .. . ... . .. ... . .. ... .. .. ... ... ... .. .. ... ... .. .. ... ... .. ... . .. . .. .") => "HELLO WORLD"
``` | algorithms | tap_code = "ABC,DEF,GHI|JKL,MNO,PQR|STU,VWX,YZ "
table = {
char: ' ' . join(['.' * k, '.' * j, '.' * i])
for i, face in enumerate(tap_code . split('|'), 1)
for j, line in enumerate(face . split(','), 1)
for k, char in enumerate(line, 1)
}
reverse = {v: k for k, v in table . items()}
def encode(s):
return ' ' . join(map(table . get, s))
def decode(s):
return '' . join(map(reverse . get, map(' ' . join, zip(* [iter(s . split())] * 3))))
| Cubic Tap Code | 62d1eb93e5994c003156b2ae | [
"Ciphers",
"Algorithms",
"Cryptography"
] | https://www.codewars.com/kata/62d1eb93e5994c003156b2ae | 6 kyu |
# Background
A high flying bird is able to estimate the contours of the ground below.
He sees hills and valleys, he sees plains and mountains.
(But you already know all this because you've solved my [Bird Mountain](https://www.codewars.com/kata/bird-mountain) Kata)
---
But this time our protagonist bird also sees a RIVER.
Not only that, but he sees that the river is rising rapidly, so much so that in a few days it threatens to inundate the surrounding land.
And all this is very important to the bird because he cannot land on water!
# Kata Task
The bird quickly calculates how much ground will remain dry as the water rises.
Can you?
## Output
* A list of how many dry landing spots there are for the **next 3 days only** (include day 0)
## Notes
* The ground is always supplied as a rectangular grid
* The normal river is at level -0.5. It is rising at 1 unit per day
* Sometimes there isn't any river at all, so you better account for those cases too!
# Example
## The birds-eye view
<table>
<tbody><tr><td>
<pre style="background:black;font-size:15px;line-height:15px;"> <span style="color:green">^^^^^^</span>
<span style="color:green">^^^^^^^^</span> <span style="color:green">^^^</span>
<span style="color:green">^^^^^^^</span> <span style="color:green">^^^</span>
<span style="color:green">^^^^^^^</span> <span style="color:green">^^^</span>
<span style="color:green">^^^^^^^</span> <span style="color:green">^^^</span>
<span style="background:blue">---------------------</span>
<span style="color:green">^^^^^</span>
<span style="color:green">^^^^^^^^</span> <span style="color:green">^^^^^^^</span>
<span style="color:green">^^^^^^^^</span> <span style="color:green">^</span> <span style="color:green">^</span>
<span style="color:green">^^^^^</span> <span style="color:green">^^^^^^^</span>
</pre>
</td>
<td width="99%">
</td></tr>
</tbody></table>
## The bird-brain calculations
<style>
#mytable, #mytable tr, #mytable tr th, #mytable tr td {
border: 1px solid orange;
border-collapse: collapse;
}
#mytable tr th {
color : orange;
}
}
</style>
<table id="mytable">
<tbody><tr><th>Day</th><th width="1%">Prediction</th><th>Number of dry landing spots
</th></tr><tr>
<td style="font-size:2em">Day 0
</td><td>
<pre style="background:black;font-size:15px;line-height:15px;"> 111111
11222221 111
1233321 111
1222221 121
1111111 111
<span style="background:blue">---------------------</span>
11111
12111111 1111111
11122111 1 1
11111 1111111
</pre>
</td><td style="font-size:4em">189
</td></tr>
<tr>
<td style="font-size:2em">Day 1
</td><td>
<pre style="background:black;font-size:15px;line-height:15px;"> 111111<span style="background:blue">-------------</span>
11222221<span style="background:blue">-------</span>111<span style="background:blue">---</span>
1233321<span style="background:blue">--</span>111<span style="background:blue">---------</span>
1222221<span style="background:blue">--</span>121<span style="background:blue">---------</span>
1111111<span style="background:blue">--</span>111<span style="background:blue">---------</span>
<span style="background:blue">---------------------</span>
11111<span style="background:blue">----------------</span>
12111111<span style="background:blue">--</span>1111111<span style="background:blue">-</span>
11122111<span style="background:blue">-----</span>1 1<span style="background:blue">-</span>
11111<span style="background:blue">--------</span>1111111<span style="background:blue">-</span>
</pre>
</td><td style="font-size:4em">99
</td></tr>
<tr>
<td style="font-size:2em">Day 2
</td><td>
<pre style="background:black;font-size:15px;line-height:15px;"><span style="background:blue">---------------------</span>
<span style="background:blue">--</span>22222<span style="background:blue">--------------</span>
<span style="background:blue">-</span>23332<span style="background:blue">---------------</span>
<span style="background:blue">-</span>22222<span style="background:blue">----</span>2<span style="background:blue">----------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">----</span>2<span style="background:blue">----------------</span>
<span style="background:blue">---</span>22<span style="background:blue">----------------</span>
<span style="background:blue">---------------------</span>
</pre>
</td><td style="font-size:4em">19
</td></tr>
<tr>
<td style="font-size:2em">Day 3
</td><td>
<pre style="background:black;font-size:15px;line-height:15px;"><span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">--</span>333<span style="background:blue">----------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
<span style="background:blue">---------------------</span>
</pre>
</td><td style="font-size:4em">3
</td></tr>
</tbody></table>
<p>
```if-not:python
`Dry ground = [189, 99, 19, 3]`
```
```if:python
`Dry ground = (189, 99, 19, 3)`
```
# Series
* [Bird Mountain](https://www.codewars.com/kata/bird-mountain)
* Bird Mountain - the river
</p> | algorithms | def dry_ground(terrain, nDays=4):
if not terrain:
return 0, 0, 0, 0
def peak_height(mountain):
lst = [[0] * Y for _ in range(X)]
for x, row in enumerate(mountain):
for y, v in enumerate(row):
here = v == '^'
isBorderOrNoPeak = not (0 < x < X - 1 and 0 < y < Y - 1 and here)
lst[x][y] = here if isBorderOrNoPeak else here + \
min(lst[x - 1][y], lst[x][y - 1])
for x in reversed(range(1, X - 1)):
for y in reversed(range(1, Y - 1)):
lst[x][y] = min(lst[x][y], lst[x + 1][y] + 1, lst[x][y + 1] + 1)
return lst
def flood(h):
bag = set(river)
while h and bag:
bag = {(a, b) for x, y in bag for a, b in ((x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1))
if (a, b) not in river and 0 <= a < X and 0 <= b < Y and flooded[a][b] < h}
river . update(bag)
return X * Y - len(river)
X, Y = len(terrain), len(terrain[0])
flooded = peak_height(terrain)
river = {(x, y) for x, r in enumerate(terrain)
for y, c in enumerate(r) if c == '-'}
return tuple(map(flood, range(nDays)))
| Bird Mountain - the river | 5c2fd9188e358f301f5f7a7b | [
"Algorithms",
"Graph Theory"
] | https://www.codewars.com/kata/5c2fd9188e358f301f5f7a7b | 4 kyu |
Image processing is a broad term that tends to be used to group algorithms that work on bidimensional arrays. This definition is often generalized for more dimensions, but we are going to keep it simple for this kata.
A binary image is an image that only has two possible colors on each pixel (normally, white and black). Here is an example with 4x10 pixels:
```
+----------+
| *** ** * |
| ****** **|
| ** * * |
| * |
+----------+
```
With this representation, the asterisks are counted as "set" pixels (white) and the spaces are counted as "empty" pixels (black).
We also say that pixels are 4-connected when they are exactly to the left, right, up or down to each other. We do not count diagonally connected pixels because that is called 8-connectivity. A 4-connected component is a set of 4-connected points that are all either set or empty. The example above has two white and four black 4-connected components.
Finally, you task is to make an algorithm that counts the number of white 4-connected components in an image given as a string in the format above (including the borders). The correct answer for the example above would be 2.
Note that the input image's dimensions are **unbounded**!
**Tip:** do not worry about coordinates or labeling. This kata is devised to be given short and elegant solutions, not efficient ones. Also, the correct string format is guaranteed.
**Python note:** images are defined as follows:
```python
img = """
+----...----+
| |
. .
. IMAGE .
. .
| |
+----...----+
"""
```
So watch out for line jumps! | algorithms | # Approach: A breadth first search (BSF) is applied to the image, where a desired cell is '*'
# The BFS is itterative
import queue
class ConnectedChecker (object):
__WHITE_CELL__ = '*'
# part of the logic to find all connected shapes
@ staticmethod
def __inImage__(i, j, nR, nC):
if 0 <= i and i < nR and 0 <= j and j < nC:
return True
else:
return False
# -----end function
# part of the logic to find all connected shapes
# A desired cell is one that is a ship symbol and has not been visited yet
@ staticmethod
def __isDesiredCell__(img, hasVisited, i, j):
if img[i][j] == ConnectedChecker . __WHITE_CELL__ and not hasVisited[i][j]:
return True
else:
return False
# ---end function
# part of the logic to find all connected shapes
@ staticmethod
def __markCellAsVisited__(hasVisited, i, j):
hasVisited[i][j] = True
# end function
# Returns a list of coordinates of the shape connected at iRoot, jRoot
@ staticmethod
def __getConnectedShape__(img, hasVisited, nR, nC, iRoot, jRoot):
# di and dj specify the directions we are going to search dx=0 dy=1 means
# search north (one cell above), di=1 dj=-1 is one to right and one down
di = [0, 0, 1, - 1]
dj = [1, - 1, 0, 0]
numSearchDirs = len(di)
# i,j are lists that are going to hold the nodes conneced to the provided root (i,j)
iConnected, jConnected = [], []
q = queue . Queue()
# if provided root is desired, save it
if ConnectedChecker . __isDesiredCell__(img, hasVisited, iRoot, jRoot):
q . put((iRoot, jRoot))
ConnectedChecker . __markCellAsVisited__(hasVisited, iRoot, jRoot)
# Now we search for all connections along the allowed (four) directions
while q . empty() == False:
u, v = q . get()
iConnected . append(u)
jConnected . append(v)
# from the current position (u,v), search in the four cardinal directions
for s in range(numSearchDirs):
iNew = u + di[s]
jNew = v + dj[s]
# if we're in the image and found a desired cell, add it to the queue
if ConnectedChecker . __inImage__(iNew, jNew, nR, nC) and \
ConnectedChecker . __isDesiredCell__(img, hasVisited, iNew, jNew):
ConnectedChecker . __markCellAsVisited__(hasVisited, iNew, jNew)
q . put((iNew, jNew))
# Should we get to this point, there does not exist a way to an exit of the maze
# from the starting location (path to exit: list(zip(xConnected, yConnected)))
return list(zip(iConnected, jConnected))
# ---end function
@ staticmethod
def countShapes(image):
img = image . split('\n')
img = [row for row in img if row]
nC = len(img[0][:])
nR = len(img)
hasVisited = [[False for _ in range(nC)] for _ in range(nR)]
totalShapes = 0
for i in range(nR):
for j in range(nC):
if not hasVisited[i][j]:
startRow, startCol = i, j
shape = ConnectedChecker . __getConnectedShape__(
img, hasVisited, nR, nC, startRow, startCol)
if shape:
totalShapes += 1
return totalShapes
# ---end function
# end class
# driver for Kata
def connected_components(img):
c = ConnectedChecker()
return c . countShapes(img)
# end driver
| Topological Image Processing: 4-Connected Components | 5e452a0a5111c7001faa2a71 | [
"Image Processing",
"Mathematics",
"Algorithms"
] | https://www.codewars.com/kata/5e452a0a5111c7001faa2a71 | 5 kyu |
<img src="https://i.imgur.com/ta6gv1i.png?1"/>
# Kata Task
A bird flying high above a mountain range is able to estimate the height of the highest peak.
Can you?
# Example
## The birds-eye view
<pre style='background:black;font-size:20px;line-height:20px;'>
<span style='color:green'>^^^^^^</span>
<span style='color:green'>^^^^^^^^</span>
<span style='color:green'>^^^^^^^</span>
<span style='color:green'>^^^^^</span>
<span style='color:green'>^^^^^^^^^^^</span>
<span style='color:green'>^^^^^^</span>
<span style='color:green'>^^^^</span>
</pre>
## The bird-brain calculations
<pre style='background:black;font-size:20px;line-height:20px;'>
<span style='color:orange'>111111</span>
<span style='color:orange'>1</span><span style='color:green'>^^^^</span><span style='color:orange'>111</span>
<span style='color:orange'>1</span><span style='color:green'>^^^^</span><span style='color:orange'>11</span>
<span style='color:orange'>1</span><span style='color:green'>^^^</span><span style='color:orange'>1</span>
<span style='color:orange'>1</span><span style='color:green'>^^^^</span><span style='color:orange'>111111</span>
<span style='color:orange'>1</span><span style='color:green'>^^^</span><span style='color:orange'>11</span>
<span style='color:orange'>1111</span>
</pre>
<p>
<pre style='background:black;font-size:20px;line-height:20px;'>
111111
1<span style='color:orange'>2222</span>111
1<span style='color:orange'>2</span><span style='color:green'>^^</span><span style='color:orange'>2</span>11
1<span style='color:orange'>2</span><span style='color:green'>^</span><span style='color:orange'>2</span>1
1<span style='color:orange'>2</span><span style='color:green'>^^</span><span style='color:orange'>2</span>111111
1<span style='color:orange'>222</span>11
1111
</pre>
<p>
<pre style='background:black;font-size:20px;line-height:20px;'>
111111
12222111
12<span style='color:orange'>33</span>211
12<span style='color:orange'>3</span>21
12<span style='color:orange'>33</span>2111111
122211
1111
</pre>
<p>
`Height = 3`
# Series
* Bird Mountain
* [Bird Mountain - the river](https://www.codewars.com/kata/bird-mountain-the-river)
| reference | def peak_height(mountain):
M = {(r, c) for r, l in enumerate(mountain)
for c in range(len(l)) if l[c] == '^'}
h = 0
while M:
M -= {(r, c)
for r, c in M if {(r, c + 1), (r, c - 1), (r + 1, c), (r - 1, c)} - M}
h += 1
return h
| Bird Mountain | 5c09ccc9b48e912946000157 | [
"Algorithms",
"Graph Theory"
] | https://www.codewars.com/kata/5c09ccc9b48e912946000157 | 5 kyu |
# Number Pyramid:
Image a number pyramid starts with `1`, and the numbers increasing by `1`.
For example, when the pyramid has `5` levels:
```
01
02 03
04 05 06
07 08 09 10
11 12 13 14 15
```
_Define the `center` as the middle number of the middle row._
And the `center` in the above case is:
```
5
```
___
# Input:
You will be given a number `n`, which means how many levels the pyramid has.
`0 <= n <= 5000`
___
# Output:
You need to return the center number of the pyramid.
* When there is no center (including `0` level), return `0`.
* When there is only `1` level, return `1` as it is the `center`.
__Examples:__
```
center(0) --> 0 or the equivalent.
center(1) --> 1 or the equivalent.
center(2) --> 0 or the equivalent.
center(3) --> 0 or the equivalent.
center(4) --> 0 or the equivalent.
center(5) --> 5 or the equivalent.
```
See more examples in the sample tests.
___
# Golfing Message:
The length of your code should be less or equal to `40`.
The length of reference solution is `35`.
___
If you like this series, welcome to do [other kata](https://www.codewars.com/collections/code-golf-number-pyramid-series) of it. | games | def center(n): return ((n + 2) * n + 5) / 8 * (n % 4 == 1)
| [Code Golf] Number Pyramid Series 3 - Center | 62cf27c0bf218d8f23bd1ca1 | [
"Mathematics",
"Restricted"
] | https://www.codewars.com/kata/62cf27c0bf218d8f23bd1ca1 | 6 kyu |
Time to train your knights...
Considere an infinite chessboard... With obstacles on it...
You're provided with a start point and an end point (as tuples `(x,y)`), and a tuple of all coordinates that are blocked by obstacles. Your goal is to find the number of minimal moves required to reach the destination, coming from the start.
If the destination isn't reachable, return `None/nil`.
<br>
### _DETAILS_
* A knight moves in an "L" shape: two positions in one direction, then one position in a perpendicular direction.
* Be extremly careful about performances, there are some strict contraints to match (see below)
* It _may_ rarely happen that the random generator fails to generate starting or ending positions. If it happens, you'll get an error message about that. Just try again, and you should be fine.
<br>
### _Performances requirements_
* Your code will be tested in different situations and should be as fast as possible.
* Moreover, its performances should be independant of the direction it searches in (meaning, they should stay the same when exchanging start and destination).
* Your code will be compared to the reference implementation and you'll have to be at most 3 to 6 times slower in the timed tests (the actual factor value depends on the kind of test).
<br>
<hr>
<br>
You are provided with 2 functions (in the preloaded section) to be able to display some inputs if needed (it's already done in the fixed tests):
```python
def display(start, dest, obst, limit=((0,20),(0,20))):
""" Display function... (how surprizing!)
start: start
dest: destination
obst: data about obstacles
limit: area of the chessboard that will be printed, as ((xmin,xmax),(ymin,ymax))
(note: axis are like indexing: x>0 is vertical descendant,
y>0 is horizontal going right)
return: None
"""
def data_builder(s,print_it=True):
""" Convert a string representation of a chessboard to the set of
corresponding inputs for your "attack" function
s: string representation of the board
printIt: display the input
return: start, dest, obst
"""
```
```ruby
# Display function... (how surprizing!)
# start: start
# dest: destination
# obst: data about obstacles
# limit: area of the chessboard that will be printed, as ((xmin,xmax),(ymin,ymax))
# (note: axis are like indexing: x>0 is vertical descendant,
# y>0 is horizontal going right)
# return: nil
def display(s,e,o,xi=0,xf=20,yi=0,yf=20) ... end
# Convert a string representation of a chessboard to the set of
# corresponding inputs for your "attack" function
# s: string representation of the board
# printIt: display the input
# return: [start, dest, obst]
def data_builder(s,print_it=true) ... end
``` | algorithms | from heapq import *
from collections import defaultdict
from itertools import product
INF, STEP = float('inf'), (2 * 2 + 1) * * .5
K_MOVES = tuple((a, b) for a, b in product(
range(- 2, 3), repeat=2) if abs(a) + abs(b) == 3)
def attack(start, end, obst):
def h(a, b, xF, yF):
a, b = a - xF, b - yF
return (a * a + b * b) * * .5 / STEP
soFar = INF
qS, qE = [(0, 0, start)], [(0, 0, end)]
sS, sE = defaultdict(lambda: INF), defaultdict(lambda: INF)
steps = ((qS, sS, sE, * end), (qE, sE, sS, * start))
for t in obst:
sS[t] = sE[t] = 0
sS[start] = sE[end] = 0
while qS and qE:
for q, s, sO, xF, yF in steps:
ch, c, pos = heappop(q)
x, y = pos
if ch > soFar:
return soFar
if pos in sO:
lenPath = s[pos] + sO[pos]
if lenPath > soFar:
return soFar
soFar = lenPath
else:
c += 1
for dx, dy in K_MOVES:
a, b = p = x + dx, y + dy
prev = s[p]
if c < prev:
heappush(q, (c + h(a, b, xF, yF), c, p))
s[p] = c
return soFar if soFar != INF else None
| Knight's Attack! | 58e6d83e19af2cb8840000b5 | [
"Algorithms",
"Graph Theory"
] | https://www.codewars.com/kata/58e6d83e19af2cb8840000b5 | 3 kyu |
*<center>I hope you enjoy this kata, and if you want, there's a [more challenging version](https://www.codewars.com/kata/58e6d83e19af2cb8840000b5).</center>*
# Task
Given two positions on an **infinite** chessboard, find the **shortest path** for a **knight** moving from one to the other.
### Input
Two arguments `start` and `end` represent the start and end positions of the knight's path respectively.
### Output
A list/array contains all the positions of the shortest path.
---
# Examples
```python
knight_path((1, -1), (1, -1)) # [(1, -1)]
knight_path((0, 0), (1, 1)) # [(0, 0), (2, -1), (1, 1)]
knight_path((0, 0), (1, 2)) # [(0, 0), (1, 2)]
knight_path((0, 0), (1, 3)) # [(0, 0), (-1, 2), (1, 3)]
knight_path((1, -3), (3, 2)) # [(1, -3), (2, -1), (1, 1), (3, 2)]
knight_path((-5, -7), (8, 7)) # [(-5, -7), (-4, -5), (-2, -4), (-1, -2), (1, -1), (2, 1), (4, 2), (5, 4), (7, 5), (8, 7)]
```
---
# Note
- The start position may **duplicate** the end position.
- There are **no obstacles** on the chessboard, so the knight can go **anywhere**.
- There may be multiple shortest paths, you only need to return **one of them**.
- Find the right algorithm and pay attention to the performance.
```if:python
- Returned path should be a **list** and **nonempty**(obviously).
- All positions should be a **tuple of two ints**.
```
---
# Tests
```if:python
- `410` fixed tests, including `6` example tests, `400` mini tests and `4` big tests.
- `100` small random tests, shortest path length in `$[90, 100)$`.
- `50` medium random tests, shortest path length in `$[9\ 000, 10\ 000)$`.
- `20` big random tests, shortest path length in `$[90\ 000, 100\ 000)$`.
- `One Last Test`, shortest path length in `$[3\ 999\ 900, 4\ 000\ 000)$`.
``` | algorithms | from itertools import product
from math import atan2, pi
MOVES = [(2, 1), (1, 2), (- 1, 2), (- 2, 1),
(- 2, - 1), (- 1, - 2), (1, - 2), (2, - 1)]
KNGIHT = 5 * * .5
CLOCK = 2 * pi
def knight_path(* pnts):
(a, b), (i, j) = pnts
ang = atan2(j - b, i - a)
da, db = min(MOVES, key=smallestAngleWith(ang))
isSameSide = sameSideOfPredFactory(i - a, j - b, da, db)
movesOnSameSide = ((di, dj) for di, dj in MOVES if isSameSide(di, dj))
opp = ang + pi if ang < 0 else ang - pi
di, dj = min(movesOnSameSide, key=smallestAngleWith(opp))
x, y = crossingPoint(a, b, da, db, i, j, di, dj)
n1, n2 = getSteps(x, y, a, b), getSteps(x, y, i, j)
path1 = nStepsWithDirFrom(n1, da, db, a, b)
path2 = nStepsWithDirFrom(n2, di, dj, i, j)
link = pathsLinker(path1[- 2:], path2[- 2:])
path = behead(path1, link[0])
path . extend(link)
path . extend(reversed(behead(path2, link[- 1])))
return path
def smallestAngleWith(ang):
def mapper(dxy):
other = atan2(dxy[1], dxy[0])
v = (other - ang) % CLOCK
return min(abs(v), abs(CLOCK - v))
return mapper
def sameSideOfPredFactory(a, b, x, y):
ref = a * y - b * x
cmp = (ref > 0) - (ref < 0)
def predicate(i, j):
v = a * j - b * i
return cmp == (v > 0) - (v < 0)
return predicate
def crossingPoint(a, b, da, db, i, j, di, dj):
if (da, db) == (- di, - dj):
return i, j # straight line => build one shot to the end...
v1 = b - a * db / da
v2 = j - i * dj / di
x = (v2 - v1) * da * di / (db * di - dj * da)
y = x * db / da + v1
return x, y
def getSteps(x, y, a, b):
return int(((a - x) * * 2 + (b - y) * * 2) * * .5 / KNGIHT)
def nStepsWithDirFrom(n, di, dj, i, j):
return [(i + di * z, j + dj * z) for z in range(n + 1)]
def behead(path, p):
while path[- 1] != p:
path . pop()
path . pop()
return path
def cachedLinks():
# all shortest paths from (0,0) to any position in a "range(-8,9)" around (x and y)
shortestTo = {(0, 0): [(0, 0)]}
bag = {(0, 0)}
while bag:
tmp = set()
for pos in bag:
for dx, dy in MOVES:
neigh = a, b = pos[0] + dx, pos[1] + dy
if - 9 < a < 9 and - 9 < b < 9 and neigh not in shortestTo:
shortestTo[neigh] = shortestTo[pos] + [neigh]
tmp . add(neigh)
bag = tmp
def getLinkFor(pair):
return shortestTo[tuple(x - a for a, x in zip(* pair))]
def pathHeadsLinker(head1, head2):
(a, b), _ = bestPair = min(product(head1, head2),
key=lambda pair: len(getLinkFor(pair)))
return [(a + x, b + y) for x, y in getLinkFor(bestPair)]
return pathHeadsLinker
pathsLinker = cachedLinks()
"""
Two to three times faster than the usual approaches.
Here is why:
When linking A to B, most soluitons try to stay as close as possible from the straight
line between the two points. But once you got a shortest path, any permutation of its
moves is also a shortest path.
So, this shortest path (on the left) with shenanigans around B to get the shortest path,
can be rearranged this way (on the right):
A A
\ /
/ /
/ */
/ */*
\ *\
/ \
\* \
*B* B
*
So, here you go:
1) Identify the two optimal directions to use
2) Find the crossing point of the two lines and build straight paths up to that point
3) Take the last 2 points on each straight path and find what combination of 2 of them
giving you the shortest path.
Those "links" are computed once and cached, so that step is O(1) as well.
4) Assemble all the parts, "beheading" the redundant/useless points of the two straight lines
Impact on performances:
- 99% of the path is built with 2 list comprehensions (which are by assence faster than
appending in a loop)
- There is no decision to be done at each step (turning right or left), here. The only
decisions to make are right at the start, to determine the 2 optimal directions to use.
This aspect is critical about the performances because it makes the step by step approach
O(N*x), with x a small constant factor (picking the right direction at each step), while
the above approach is in comparison like O(N+x), because the optimal directions are picked
once at the beginning.
Both are linear in time, but the constant factor is lower here.
"""
| Shortest Knight Path(Infinite Chessboard) | 62287e1766b26a0024b9e806 | [
"Performance",
"Algorithms",
"Graph Theory"
] | https://www.codewars.com/kata/62287e1766b26a0024b9e806 | 4 kyu |
Your friend has a list of ```k``` numbers : ```[a1, a2, a3, ... ak]```.
He is allowed to do the operation which consists of three steps:
1) select two numbers: ```ai``` and ```aj``` (```ai % 2 = 0```)
2) replace ```ai``` with ```ai``` / 2
3) replace ```aj``` with ```aj``` * 2
Help him to find the maximum sum of list elements that is possible to achieve by using this operation(maybe multiple times). Return this sum modulo 1 000 000 007, because it can be quite big.
<b>Input:</b>
List of k elements: [a1, a2, a3, ..., ak]; k < 10**4
All numbers are positive and smaller than 10**9
<b>Output:</b>
Maximum sum after some operations(modulo 1 000 000 007)
| games | def divide_and_multiply(n):
s, d = 1, []
for x in n:
b = x & - x
d . append(x / / b)
s *= b
return (sum(d) + max(d) * (s - 1)) % (10 * * 9 + 7)
| Divide and maximize | 62275b5bf6c169002379fa65 | [
"Puzzles",
"Lists",
"Arrays",
"Mathematics"
] | https://www.codewars.com/kata/62275b5bf6c169002379fa65 | 5 kyu |
### Story
There are 100 prisoners in solitary cells. There's a central living room with one light bulb; this bulb is initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner at random, and that prisoner visits the living room. While there, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world could always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity. The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion?
### Task
Change the ```living_room``` function in the initial code to allow the prisoners to win their freedom in every game.
The function ```living_room``` takes in three argument, the first of which is an integer that identifies a prisoner. To call ```living_room``` with ```i``` as first argument means to call the prisoner ```#i``` into the living room.
The function returns a tuple (array in JavaScript) of two booleans, the latter of which represents the assertion. If the second value of the tuple returned is ```True``` an assertion is made by the prisoner in the living room.
A game is a sequence of calls of the function ```living_room``` that ends when an assertion is made (or after 29200 times if no assertion is made). The prisoners win the game and are set free if an assertion is made after all the prisoners were called into the living room at least once during the game.
### Input
* an integer between 0 and 99 for the prisoner number
* a boolean for the current state of the light bulb
* an array of booleans that represents the states of the light bulb during the previous visits of the prisoner in the room (when they go in, before changing it)
### Output
A tuple (array in JavaScript) with:
* a boolean for the new state of the light bulb
* a boolean for the assertion
### Example
In the example test the prisoners are taken in the room in order, from first to last, then the cycle repeats.
For example, prisoner #0 goes in on day 1, when the lightbulb is off and they have no previous visits, and then again on day 101, finding the lightbulb in whatever state was left by prisoner #99, and with ```previous_visits == [False]```.
If an assertion is:
* made before the 100th day, the prisoners are shot;
* made at or after the 100th day, the prisoners are set free;
* not made after 80 years (29200 days), the prisoners are shot.
### Notes
To make sure that none of the prisoners is communicating, your solution should not rely on any kind of state.
The arguments to the function are designed to make keeping state unnecessary.
The only information needed by the prisoners a prisoner will have at any time is:
* the predecided plan (i.e. the ```living_room``` function)
* who they themselves are and what role they play in the plan (```prisoner_number```)
* the current state of the lightbulb (```lightbulb```), and all the previous times which they had seen the lightbulb since the game started (```previousVisits```)
The prisoners may also wish to remember conclusions they had arrived at previously, however since they have all the same information they did when they reached that conclusion, they can merely do it again.
The tests are designed to fail solutions that use a state. Please do not:
* use closures
* add attributes to the function
* add other arguments to the function
* use other variables in the global scope
* rename ```living_room```
### Constraints
* the prisoners have 80 years (29200 days) to make an assertion
* the initial state of the light bulb is off (False)
### Bonus points
Solve the task without the assumption that the light bulb is initially off for a bonus point.
The initial state of the light bulb can be on or off, the prisoners know that is chosen at random but they don't know what is the initial state. They also don't know when the game starts.
| games | # Prisoner 0 is the leader
# He always leaves the lightbulb on
# If he found the lightbulb off 100 times, it means everyone visited at least once
# Other prisoners must turn off the lightbulb only if:
# - they never turned it off before
# - it is currently on
# TLDR:
# - turning the lightbulb on means "I visited"
# - do not turn it on if you already did it
# - prisoner 0 is in charge of resetting the lightbulb
def living_room(prisoner_number, lightbulb, previous_visits):
if prisoner_number == 0:
if len(previous_visits) - sum(previous_visits) == 100:
return True, True
return True, False
if lightbulb and not any(previous_visits):
return False, False
return lightbulb, False
| 100 prisoners and a light bulb | 619d3f32156bbd000896463c | [
"Riddles",
"Puzzles"
] | https://www.codewars.com/kata/619d3f32156bbd000896463c | 5 kyu |
The national go-kart racing competition is taking place at your local town and you've been called for building the winners podium with the available wooden blocks. Thankfully you are in a wood-rich area, number of blocks are always at least 6.
Remember a classic racing podium has three platforms for first, second and third place. First place is the highest and second place is higher than third. Also notice that platforms are arranged as 2nd - 1st - 3rd.
The organizers want a podium that satisfies:
- The first place platform has the **minimum** height possible
- The second place platform has the **closest** height to first place
- All platforms have heights greater than zero.
## Task
Given the numbers of blocks available, return an array / tuple or another data structure depending on the language (refer sample tests) with the heights of 2nd, 1st, 3rd places platforms.
## Examples (input -> output)
```
11 -> [4, 5, 2]
6 -> [2, 3, 1]
10 -> [4, 5, 1]
```
| reference | from math import ceil
def race_podium(b):
f = ceil((b) / 3) + 1
s = min(f - 1, b - f - 1)
return s, f, b - f - s
| Race Ceremony | 62cecd4e5487c10028996e04 | [
"Algorithms",
"Mathematics"
] | https://www.codewars.com/kata/62cecd4e5487c10028996e04 | 7 kyu |
## Overview
Given a 2-dimensional `m * n` rectangular grid with non-negative integer coordinates, can you develop an algorithm that will place the largest possible number, `P`, of points on the grid **without forming ANY right triangles in the resulting configuration of points** ?
Language clarification: depending on your country, a *"right triangle"* can be also called *"rectangle triangle"*, *"orthogonal triangle"*, *"right-angle triangle"* etc. It is a triangle which contains one angle that equals 90 degrees.
---
## Inputs
You will be given two positive `integers`, `m >= 1` and `n >= 1`, representing the dimensions of an `m * n` rectangular grid with integer coordinates.
Values of `m` and `n` will be tested up to `50`.
## Outputs
You must return a `list` (Python) of 2-element `tuples`, where each `tuple` represents the location of a point that you want to place in the rectangular grid.
Each tuple must be of the form `(x,y)` where `x` and `y` must both be `integers`.
The order of the returned list does not matter, so no need to sort it etc.
**We will count the integer coordinates from `x = 0` and `y = 0`, so that the 4 corner points of the given grid will always be `(0,0)`, `(0,n-1)`, `(m-1,0)`, `(m-1,n-1)`**
---
## Troubleshooting and Test Explanation
From your returned solution `list`, the tests will first check that your solution is valid i.e. that it contains `integer` 2-ples whose coordinates are not outside the allowed range of `0 <= x <= m-1` and `0 <= y <= n-1`.
Then the tests will check that there are **no right triangles formed by any 3 of the points that you have placed.**
Please pay attention to the possibility of forming right triangles whose right-angle is not "aligned" by the x- and y-axes.
For example, the triangle given by the 3 points `{(1, 1), (2, 2), (1, 3)}` **is in fact** a right triangle, even though you might not notice it in your grid.
Finally, if your solution list doesn't contain any right triangles, it will then be tested for **optimality** i.e. that you have managed to fit as many points as possible into the grid.
If you fail at this last part of the tests, it is because there is a way to place even more points in the given grid than what you have found using your current algorithm.
---
## Example With NON-OPTIMAL Solution
Consider the following illustration:
<a href="https://ibb.co/r6cY4xg"><img src="https://i.ibb.co/MVGrMcb/kata-illustration.png" alt="kata-illustration" border="0"></a>
Here we are dealing with an `m=7` and `n=5` grid; the 4 black points should clarify that the coordinates therefore start with `x=0`, `y=0`.
In this example, the user has succesfully placed a total of `P = 6` points represented in red, so his solution would be returned as the list (order is not important):
`[(0,1), (1,3), (2,2), (4,2), (5,3), (6,1)]`
You can confirm for yourself manually, or by using the solution-checker that is loaded in the test cases, that so far these `P = 6` points **do not form any right triangles**.
You can also experiment with the solution-checker and try adding *any* point with x-coordinate `x = 3` to the above list; you will find that adding any such point **will** create a right triangle somewhere.
Finally, this example **with P = 6 points is not optimal for the given values of m and n**: there are ways of placing **P = 10 total points in this grid**. | reference | def most_points(m, n):
return [(0, 0)] if n == m == 1 else [(0, y) for y in range(m > 1, n)] + [(x, 0) for x in range(n > 1, m)]
| Most points without right triangle | 62c75ce198d93d02a42ca7a3 | [
"Puzzles",
"Mathematics",
"Geometry"
] | https://www.codewars.com/kata/62c75ce198d93d02a42ca7a3 | 6 kyu |
Background
==========
This kata is based on the [Magic Triangle](https://en.wikipedia.org/wiki/Magic_triangle_(mathematics)) math puzzle, invented in 1972 by math teacher Terrel Trotter, Jr.
The puzzles have only 2 constraints:
* Each of the digits 1 through 9 can be used only once.
* Every side must sum to the same number.
For example, here is an unsolved and then solved Magic Triangle puzzle:
<svg xmlns="http://www.w3.org/2000/svg" version="1.1" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:svgjs="http://svgjs.dev/svgjs" width="640" height="576" viewBox="0 0 640 576" style="height: 200px; width: auto"><line x1="320" y1="80" x2="560" y2="496" stroke-width="5" stroke="grey"></line><line x1="560" y1="496" x2="80" y2="496" stroke-width="5" stroke="grey"></line><line x1="80" y1="496" x2="320" y2="80" stroke-width="5" stroke="grey"></line><circle r="60" cx="320" cy="80" stroke="grey" stroke-width="5px" fill="LightGray"></circle><circle r="60" cx="400" cy="219" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="400" y="234" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="400">4</tspan></text><circle r="60" cx="480" cy="357" stroke="grey" stroke-width="5px" fill="LightGray"></circle><circle r="60" cx="560" cy="496" stroke="grey" stroke-width="5px" fill="LightGray"></circle><circle r="60" cx="400" cy="496" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="400" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="400">2</tspan></text><circle r="60" cx="240" cy="496" stroke="grey" stroke-width="5px" fill="LightGray"></circle><circle r="60" cx="80" cy="496" stroke="grey" stroke-width="5px" fill="LightGray"></circle><circle r="60" cx="160" cy="357" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="160" y="372" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="160">5</tspan></text><circle r="60" cx="240" cy="219" stroke="grey" stroke-width="5px" fill="LightGray"></circle></svg>
<svg xmlns="http://www.w3.org/2000/svg" version="1.1" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:svgjs="http://svgjs.dev/svgjs" width="640" height="576" viewBox="0 0 640 576" style="height: 200px; width: auto"><line x1="320" y1="80" x2="560" y2="496" stroke-width="5" stroke="grey"></line><line x1="560" y1="496" x2="80" y2="496" stroke-width="5" stroke="grey"></line><line x1="80" y1="496" x2="320" y2="80" stroke-width="5" stroke="grey"></line><circle r="60" cx="320" cy="80" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="320" y="95" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="320">9</tspan></text><circle r="60" cx="400" cy="219" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="400" y="234" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="400">4</tspan></text><circle r="60" cx="480" cy="357" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="480" y="372" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="480">3</tspan></text><circle r="60" cx="560" cy="496" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="560" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="560">7</tspan></text><circle r="60" cx="400" cy="496" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="400" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="400">2</tspan></text><circle r="60" cx="240" cy="496" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="240" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="240">6</tspan></text><circle r="60" cx="80" cy="496" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="80" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="80">8</tspan></text><circle r="60" cx="160" cy="357" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="160" y="372" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="160">5</tspan></text><circle r="60" cx="240" cy="219" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="240" y="234" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="240">1</tspan></text></svg>
It should be noted that a Magic Triangle may have more than one solution. For example, the above puzzle can also be solved like this:
<svg xmlns="http://www.w3.org/2000/svg" version="1.1" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:svgjs="http://svgjs.dev/svgjs" width="640" height="576" viewBox="0 0 640 576" style="height: 200px; width: auto"><line x1="320" y1="80" x2="560" y2="496" stroke-width="5" stroke="grey"></line><line x1="560" y1="496" x2="80" y2="496" stroke-width="5" stroke="grey"></line><line x1="80" y1="496" x2="320" y2="80" stroke-width="5" stroke="grey"></line><circle r="60" cx="320" cy="80" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="320" y="95" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="320">6</tspan></text><circle r="60" cx="400" cy="219" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="400" y="234" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="400">4</tspan></text><circle r="60" cx="480" cy="357" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="480" y="372" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="480">8</tspan></text><circle r="60" cx="560" cy="496" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="560" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="560">3</tspan></text><circle r="60" cx="400" cy="496" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="400" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="400">2</tspan></text><circle r="60" cx="240" cy="496" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="240" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="240">7</tspan></text><circle r="60" cx="80" cy="496" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="80" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="80">9</tspan></text><circle r="60" cx="160" cy="357" stroke="grey" stroke-width="5px" fill="yellow"></circle><text x="160" y="372" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="160">5</tspan></text><circle r="60" cx="240" cy="219" stroke="grey" stroke-width="5px" fill="lightgreen"></circle><text x="240" y="234" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="sans-serif" font-weight="bold"><tspan dy="0" x="240">1</tspan></text></svg>
Your Input
==========
Your function, `magic_triangle_solutions()` will recieve a list of numbers which will represent a Magic Triangle by index, as follows:
<svg xmlns="http://www.w3.org/2000/svg" version="1.1" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:svgjs="http://svgjs.dev/svgjs" width="640" height="576" viewBox="0 0 640 576" style="height: 200px; width: auto"><line x1="320" y1="80" x2="560" y2="496" stroke-width="5" stroke="grey"></line><line x1="560" y1="496" x2="80" y2="496" stroke-width="5" stroke="grey"></line><line x1="80" y1="496" x2="320" y2="80" stroke-width="5" stroke="grey"></line><circle r="60" cx="320" cy="80" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="320" y="95" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="320">[0]</tspan></text><circle r="60" cx="400" cy="219" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="400" y="234" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="400">[1]</tspan></text><circle r="60" cx="480" cy="357" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="480" y="372" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="480">[2]</tspan></text><circle r="60" cx="560" cy="496" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="560" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="560">[3]</tspan></text><circle r="60" cx="400" cy="496" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="400" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="400">[4]</tspan></text><circle r="60" cx="240" cy="496" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="240" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="240">[5]</tspan></text><circle r="60" cx="80" cy="496" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="80" y="511" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="80">[6]</tspan></text><circle r="60" cx="160" cy="357" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="160" y="372" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="160">[7]</tspan></text><circle r="60" cx="240" cy="219" stroke="grey" stroke-width="5px" fill="LightGray"></circle><text x="240" y="234" dominant-baseline="middle" text-anchor="middle" font-size="60px" font-family="monospace" font-weight="bold"><tspan dy="0" x="240">[8]</tspan></text></svg>
The digits 1 through 9 will be represented by their integers, and a blank space by the integer 0. Thus, the unsolved puzzle from the previous section would be represented by the following list:
`[0, 4, 0, 0, 2, 0, 0, 5, 0]`
Your input lists will only contain integers of the range 0 through 9, with no non-zero numbers repeated, so you won't have to worry about validating your inputs.
Your Output
===========
Your function will output a list containing all possible solutions to the provided puzzle.
For example, `magic_triangle_solutions([0, 4, 0, 0, 2, 0, 0, 5, 0])` should return
```
[
[6, 4, 8, 3, 2, 7, 9, 5, 1],
[8, 4, 6, 3, 2, 9, 7, 5, 1],
[9, 4, 3, 7, 2, 6, 8, 5, 1]
]
```
You won't have to worry about the order of solutions in your return list. For example,
```
[
[9, 4, 3, 7, 2, 6, 8, 5, 1],
[8, 4, 6, 3, 2, 9, 7, 5, 1],
[6, 4, 8, 3, 2, 7, 9, 5, 1]
]
```
would also be an acceptable return value.
Note
====
There will be 100 randomized tests and a fully blank list to solve in the full test suite, so your code must be somewhat performant to pass. However, the timeout length is such that brute force testing should still pass provided that no other performance issues get in the way.
Good luck, and enjoy!
---------------------
| games | from itertools import permutations
# Missing digits
def get_unknown(puzzle):
return set(range(1, 10)) - set(puzzle)
# All possible tries to solve the puzzle
def get_possible(puzzle):
for perm in map(iter, permutations(get_unknown(puzzle))):
yield [x or next(perm) for x in puzzle]
# Check if a solution is valid
def check_possible(result):
return sum(result[: 4]) == sum(result[3: 7]) == sum(result[6:]) + result[0]
# Main function
def magic_triangle_solutions(puzzle):
return [* filter(check_possible, get_possible(puzzle))]
| Magic Triangle Solutions | 62b2072d62c66500159693ff | [
"Arrays",
"Algorithms",
"Game Solvers",
"Games",
"Mathematics"
] | https://www.codewars.com/kata/62b2072d62c66500159693ff | 6 kyu |
You are given three integer inputs: length, minimum, and maximum.
Return a string that:
1. Starts at minimum
2. Ascends one at a time until reaching the maximum, then
3. Descends one at a time until reaching the minimum
4. repeat until the string is the appropriate length
Examples:
```
length: 5, minimum: 1, maximum: 3 ==> "12321"
length: 14, minimum: 0, maximum: 2 ==> "01210121012101"
length: 11, minimum: 5, maximum: 9 ==> "56789876567"
```
Notes:
- length will always be non-negative
- negative numbers can appear for minimum and maximum values
- hyphens/dashes ("-") for negative numbers do count towards the length
- the resulting string must be truncated to the exact length provided
- return an empty string if maximum < minimum or length == 0
- minimum and maximum can equal one another and result in a single number repeated for the length of the string
| algorithms | def ascend_descend(l, m, M): return '' . join(
map(str, l * [* range(m, M + 1), * range(M - 1, m, - 1)]))[: l]
| Ascend, Descend, Repeat? | 62ca07aaedc75c88fb95ee2f | [
"Fundamentals",
"Strings",
"Algorithms"
] | https://www.codewars.com/kata/62ca07aaedc75c88fb95ee2f | 6 kyu |
## Description
An infinite number of shelves are arranged one above the other in a staggered fashion.<br>
The cat can jump either one or three shelves at a time: from shelf `i` to shelf `i+1` or `i+3` (the cat cannot climb on the shelf directly above its head), according to the illustration:
```
┌────────┐
│-6------│
└────────┘
┌────────┐
│------5-│
└────────┘ ┌─────► OK!
│ ┌────────┐
│ │-4------│
│ └────────┘
┌────────┐ │
│------3-│ │
BANG!────┘ ├─────► OK!
▲ |\_/| │ ┌────────┐
│ ("^-^) │ │-2------│
│ ) ( │ └────────┘
┌─┴─┴───┴┬──┘
│------1-│
└────────┘
```
## Input
Start and finish shelf numbers (always positive integers, finish no smaller than start)
## Task
Find the minimum number of jumps to go from start to finish
## Example
Start ```1```, finish ```5```, then answer is ```2``` ```(1 => 4 => 5 or 1 => 2 => 5)```
## Inspirers
| algorithms | def solution(start, finish):
return sum(divmod(finish - start, 3))
| Cats and shelves | 62c93765cef6f10030dfa92b | [
"Algorithms"
] | https://www.codewars.com/kata/62c93765cef6f10030dfa92b | 7 kyu |
# Description
---
The term "google dorking" describes the process of using
filters (also called "operators") in google search
queries which limit the search results according to the
used filters. It's a technique often used by "hackers"
in order to find valuable information about a target.
*But thats not what we do here*
The syntax of most filters looks as following: `<filter>:<value>`
Hence, a complete search query could look like
```
intext:kyu site:codewars.com
```
If you enter the above query into Googles search bar, your results
are limited to **codewars.com** and every result page should
contain the word **kyu**.
*Pretty useful, huh?*
---
## Task
A coder friend of yours gave you some queries which -he promises- will
return interesting results. But some of the queries contain outdated
filters which don't work anymore. Hence, you decide to code a function
named `is_valid` (`isValid` for js) which takes in a search query of
type `str` and validates that every filter within the query is up-to-date.
If only one filter is invalid, then the entire query is invalid.
---
## Preloaded
~~~if:python
`FILTERS` - A tuple of **valid** search filters
~~~
~~~if:javascript
`FILTERS` - An array of **valid** search filters
~~~
---
## Note
- A search query will contain `0 < n < 100` filters
~~~if:python
- Your function should return `True` for valid and `False` for invalid
~~~
~~~if-not:python
- Your function should return `true` for valid and `false` for invalid
~~~
- Each filter follows the `<filter-name>:<value>` syntax
- Filters are seperated by a space (\s)
- You only need to check if the filter-name is valid
- For a filter-name to be valid, the FILTERS array must contain this filter-name.
---
## Fun Fact
```
intext:"google dorking" site:codewars.com
```
Entering the above query into Googles search bar will give you a link to this kata! (this link used to be the only result of this query until the approval of this kata, since then the query has also showed pages that have this kata in the ```similar katas``` section)
---
*Good Luck :)* | reference | FILTERS = set(FILTERS)
def is_valid(query: str) - > bool:
return all(u . split(':')[0] in FILTERS for u in query . split() if ':' in u)
| Google Dorking - Validating Queries | 62cb487e43b37a5829ab5752 | [
"Regular Expressions"
] | https://www.codewars.com/kata/62cb487e43b37a5829ab5752 | 7 kyu |
_NOTE: It is recommended to complete the [corner case](https://www.codewars.com/kata/62c4ad0e86f0166ec7bb8485) before working on this `edge case`._
___
# Number Pyramid:
Image a number pyramid starts with `1`, and the numbers increasing by `1`.
For example, when the pyramid has 4 levels:
```
1
2 3
4 5 6
7 8 9 10
```
And the sum of all numbers of the edges (including corners) are:
```
1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 10 = 50 (5 is not on the edges)
```
___
# Input:
You will be given a number `n`, which means how many levels the pyramid has.
`0 <= n <= 5000`
___
# Output:
You need to return the sum of all numbers on the edges (corners are included) of the pyramid.
* When there is no level, return `0` as there is no edge.
* When there is only one level, return `1` as it is the only number on edge/corner.
__Examples:__
```
sum_edges(0) --> 0 or 0.0
sum_edges(1) --> 1 or 1.0
sum_edges(2) --> 6 or 6.0
sum_edges(3) --> 21 or 21.0
sum_edges(4) --> 50 or 50.0
```
___
# Golfing Message:
The length of your code should be less or equal to `50`.
It is possible to do it within the length of `46`.
___
If you like this series, welcome to do [other kata](https://www.codewars.com/collections/code-golf-number-pyramid-series) of it. | games | def sum_edges(n): return (n, n * (5 * n * n - 3 * n + 10) / 6 - 2)[n > 1]
| [Code Golf] Number Pyramid Series 2 - Sum of Edges | 62c74a39fba7810016e601b1 | [
"Mathematics",
"Restricted"
] | https://www.codewars.com/kata/62c74a39fba7810016e601b1 | 6 kyu |
# Sexagenary Cycle
In trandition East Asia calendar, [Sexagenary Cycle](https://en.wikipedia.org/wiki/Sexagenary_cycle) was used to note the years. It has 2 parts:
[Heavenly Stems](https://en.wikipedia.org/wiki/Heavenly_Stems), (天干)
and
[Earthly Branches](https://en.wikipedia.org/wiki/Earthly_Branches), (地支)
The Heavenly Stems contains `10` characters:
```
甲乙丙丁戊己庚辛壬癸
```
The Earthly Branches contains `12` characters:
```
子丑寅卯辰巳午未申酉戌亥
```
__NONE__ of the characters is repeated.
___
# Year
Let's start with the beginning, year `1`, the characters for it will be:
```
01: 甲子 (Heavenly Stems always comes before Earthly Branches)
```
Both are the first character.
And `1` year goes by, now both Heavenly Stems and Earthly Branches will change to the next one, which is:
```
02 :乙丑
```
And keep going on for the following years.
```
03: 丙寅, 04: 丁卯, ...
```
And when it reaches the last character of _Heavenly Stems_ or _Earthly Branches_, it will go back to the beginning of it:
```
10: 癸酉 => 11: 甲戌 (癸 is the last character, so the next one it goes back to the first character, which is 甲)
```
And after `60` years, it will go back to beginning.
```
60: 癸亥 => 01: 甲子
```
___
# Task
You will be given 2 strings of Sexagenary Cycle, first one is the `birth year` and second one is the `present year`. You need to return how old the person is.
All inputs will be valid.
`1 <= age <= 60`
Examples **(Input1, Input2) --> Output**:
```
("甲子", "乙丑") --> 1
("甲子", "甲子") --> 60
("乙丑", "甲子") --> 59
``` | reference | HEAVENLY_STEMS = '甲乙丙丁戊己庚辛壬癸'
EARTHLY_BRANCHES = '子丑寅卯辰巳午未申酉戌亥'
YEAR = dict(
zip(map('' . join, zip(HEAVENLY_STEMS * 6, EARTHLY_BRANCHES * 5)), range(60)))
def how_old(birth_year, present_year):
a, b = YEAR[birth_year], YEAR[present_year]
return b - a + 60 * (a >= b)
| Sexagenary Cycle (干支): How old am I? | 62cc917fedc75c95ef961ad1 | [
"Fundamentals"
] | https://www.codewars.com/kata/62cc917fedc75c95ef961ad1 | 7 kyu |
You're most likely familiar with Pascal's Triangle. If you aren't, I recommend trying [the kata found here](https://www.codewars.com/kata/52945ce49bb38560fe0001d9) on which this kata is based. The 2-D Pascal's triangle can be generalized to 3-D in the form of Pascal's Pyramid (technically a tetrahedron).
Each layer of the pyramid is a triangle with the following form. Each side of a given layer is one cell longer than the sides of the previous layer. All three corner cells are always 1. All other cells are equal to the sum of the 3 (or 2 for cells on the edge of the new layer) closest cells in the previous layer. For example, below are the first 5 layers of the pyramid.
```
1 1 1 1 1
1 1 2 2 3 3 4 4
--> --> 1 2 1 --> 3 6 3 --> 6 12 6
1 3 3 1 4 12 12 4
1 4 6 4 1
n=1 n=2 n=3 n=4 n=5
```
Your task is to define a function `pascal_pyr_layer` that provides the `n`-th layer of Pascal's Pyramid.
## Closest Cells in the Previous Layer
This section clarifies what cells in a higher layer are used to find the value of a cell in the next lower layer. As an example, take layers 3 -> 4:
```
1 1
2 2 -> 3 3
1 2 1 3 6 3
1 3 3 1
```
If you overlay these on top of eachother and look at them from above, the elements in the higher layer (layer 3 here) will be located just below (from our perspective) the elements with the same position in the lower layer. For instance, the two layers above can be overlain as so (level 3 in italics, level 4 in bold):
<pre>
<b>1</b>
<i>1</i>
<b>3</b> <b>3</b>
<i>2</i> <i>2</i>
<b>3</b> <b>6</b> <b>3</b>
<i>1</i> <i>2</i> <i>1</i>
<b>1</b> <b>3</b> <b>3</b> <b>1</b>
</pre>
From this, we can see that every element in layer 4 is the sum of the (one to three) elements in layer 3 that are immediately in the down, up-right, and up-left directions from a top-down view. These are also the three closest elements in the layer above.
## Input
An integer `n` in the range `1 <= n <= 100`
## Output
The `n`-th layer of the pyramid. This should be formatted as a nested array. For example, the 5-th layer of the pyramid
```
1
4 4
6 12 6
4 12 12 4
1 4 6 4 1
n=5
```
should be returned as `[[1], [4, 4], [6, 12, 6], [4, 12, 12, 4], [1, 4, 6, 4, 1]]` | algorithms | from math import factorial
from functools import lru_cache
f = lru_cache(maxsize=None)(factorial)
def pascal_pyr_layer(n):
return [[f(n - 1) / / f(n - 1 - i) / / f(i - j) / / f(j) for j in range(i + 1)] for i in range(n)]
| Levels of Pascal's Pyramid | 62c7b2ebb8d7c8005e3b1f23 | [
"Arrays",
"Algorithms",
"Mathematics",
"Combinatorics",
"Recursion"
] | https://www.codewars.com/kata/62c7b2ebb8d7c8005e3b1f23 | 6 kyu |
# Async Request Manager
---
Your task is it to implement a coroutine which
sends `n` requests and concats the returned
strings of each response to one string.
A request can be made by invoking the prebuild
coro `send_request`. When invoked, the coro
sleeps for 1 sec and then returns a single
character.
Your overall task is to implement an async
`request_manager` which completes all `n` requests
in under **1.5** sec
---
## Summary:
- Implement a coro `request_manager` which sends and processes fake requests
- It accepts the parameter `n` which represents the amount of requests that should be made
- Requests have to be made with `send_request`
- `send_request` is a preloaded coroutine : `(coro) send_request() -> str`
- It takes 1 sec until a character is returned to the caller
- The waiting time is implemented via `asyncio.sleep`
- Characters returned by `send_request` have to be concat to one string
- That string is then returned by your `request_manager`
- The characters in the string need to be ordered, so that the character returned
by the first request is at index 0, the character returned by the second request
is at index 1 etc.
- Your `request_manager` has 1.5 sec to start & process all `n` requests and to
return the desired string
---
Good Luck, should be easy :) | reference | import asyncio
async def request_manager(n: int) - > str:
return '' . join(await asyncio . gather(* (send_request() for _ in range(n))))
| Async Requests | 62c702489012c30017ded374 | [
"Asynchronous"
] | https://www.codewars.com/kata/62c702489012c30017ded374 | 6 kyu |
# Number Pyramid
Image a number pyramid starts with `1`, and the numbers increasing by `1`.
For example, when the pyramid has 3 levels:
```
1
2 3
4 5 6
```
And the sum of three corners are:
```
1 + 4 + 6 = 11
```
___
# Input:
You will be given a number `n`, which means how many levels the pyramid has.
`0 <= n <= 5000`
___
# Output:
You need to return the sum of three corners of the pyramid.
* When there is no level, return `0` as there is no corner.
* When there is only one level, return `1` as it is the only corner.
__Examples:__
```
sum_corners(0) --> 0
sum_corners(1) --> 1
sum_corners(2) --> 6
sum_corners(3) --> 11
```
___
# Golfing Message:
The length of your code should be less or equal to `35`.
___
If you like this series, welcome to do [other kata](https://www.codewars.com/collections/code-golf-number-pyramid-series) of it. | games | def sum_corners(n): return n * n + 2 * (n > 1)
| [Code Golf] Number Pyramid Series 1 - Sum of Corners | 62c4ad0e86f0166ec7bb8485 | [
"Mathematics",
"Restricted"
] | https://www.codewars.com/kata/62c4ad0e86f0166ec7bb8485 | 7 kyu |
<h1>Task</h1>
Your task is to write a function that takes a range of *Google Sheets* cells as a parameter, and returns an array of addresses of all cells in the specified range.
<img src="https://i.ibb.co/WzVZ4PV/example.png" alt="example" border="0">
#### Notes
* Letters in addresses: from A-Z (_Google Sheets_ should not be confused with _Excel_ spreadsheets. In Excel, the _"Z"_ column is followed by _"AA"_, etc. In Google Sheets, the last column is _"Z"_)
* The final array must be sorted by the number in each address (ascending)
* The letters in the addresses must go in alphabetical order (A to Z).
For example, if the range is _"A1:C3"_, the function should return
`[ "A1", "B1", "C1", "A2", "B2", "C2", "A3", "B3", "C3" ]`, <br> not `[ "B1", "A1", "C1", "C2", "B2", "A2", "A3", "C3", "B3" ]` or something else.
* If an invalid range is passed to the function, the function should return `[]`.
The ranges that are considered to be invalid:
1. Those in which the first cell is further away than the second.
Example: `H7:F3`
2. Those in which two identical cells are specified.
Example: `C2:C2`
In Google Sheets you cannot find such a range, because it is considered as one cell. For this reason, such a range is invalid in this kata.
<br>
<h1>More examples</h1>
`"B1:H5"`:
```javascript
[
"B1", "C1", "D1", "E1", "F1", "G1", "H1",
"B2", "C2", "D2", "E2", "F2", "G2", "H2",
"B3", "C3", "D3", "E3", "F3", "G3", "H3",
"B4", "C4", "D4", "E4", "F4", "G4", "H4",
"B5", "C5", "D5", "E5", "F5", "G5", "H5"
]
```
`"A2:B3"`:
```javascript
[ "A2", "B2", "A3", "B3" ]
```
`"A1:A5"`:
```javascript
[ "A1", "A2", "A3", "A4", "A5" ]
```
`"X8:Z20"`:
```javascript
[
"X8", "Y8", "Z8",
"X9", "Y9", "Z9",
"X10", "Y10", "Z10",
"X11", "Y11", "Z11",
"X12", "Y12", "Z12",
"X13", "Y13", "Z13",
"X14", "Y14", "Z14",
"X15", "Y15", "Z15",
"X16", "Y16", "Z16",
"X17", "Y17", "Z17",
"X18", "Y18", "Z18",
"X19", "Y19", "Z19",
"X20", "Y20", "Z20"
]
```
`"E345:J346"`:
```javascript
[
"E345", "F345", "G345", "H345", "I345", "J345",
"E346", "F346", "G346", "H346", "I346", "J346"
]
```
<u>**Good Luck!**</u> | algorithms | def get_cell_addresses(cell_range):
x, y = cell_range . split(':')
if x == y:
return []
a, b, c, d = ord(x[0]), abs(int(x[1:])), ord(y[0]), abs(
int(y[1:])) # abs because the tests give some negative values
return [f" { chr ( i )}{ j } " for j in range(b, d + 1) for i in range(a, c + 1)]
| Get the addresses of all Google Sheets cells in the range | 62c376ce1019024820580309 | [
"Algorithms",
"Arrays"
] | https://www.codewars.com/kata/62c376ce1019024820580309 | 6 kyu |
Implement a generator that, given a pattern, returns strings following such pattern. In this kata a pattern is a string composed of characters, digits, punctuation (except square brackets) and tokens that have to be replaced by the value needed to build the next string in the sequence.<br><br>
There are four types of tokens:<br>
- ```[INC_INT=START,STEP]```: Incrementing integer, every time the next function is called on the generator the actual integer is put in the string and then incremented by STEP. The default values for START and STEP are ```1``` and ```1```;
- ```[INC_FLOAT=START,STEP]```: Incrementing floating point, every time the next function is called on the generator the actual floating point is put in the string and then incremented by STEP. The default values for START and STEP are ```0.1``` and ```0.1```. The ```[INC_FLOAT]``` token takes its floating point precision from the parameter that has the most decimal digits in it (trailing zeros count aswell). The floating point numbers passed as parameters might be very small (the lowest number possible is ```0.0000000001```);
- ```[INTERVAL=FIRST,LAST]```: Interval of integers starting from FIRST and ending at LAST (both included). When LAST is reached, the token restarts the interval from FIRST. The default values for FIRST and LAST are ```1``` and ```1```. If only one parameter is given, LAST is equal to FIRST;
- ```[PERIODIC=START,N]```: Every time N strings get created, the value is incremented by one. The default values for START and N are ```1``` and ```1```. Note that in the tests N is always >= 1.
<br><br>
As already mentioned, the tokens have default values: this means that you might find tokens with 0, 1 or 2 parameters. If only one parameter is given, it is the first one in the token's specification.<br>
## Examples:
```
INPUT: "Testing [INC_INT], [INC_INT=2], [INC_INT=3,2]"
OUTPUT: "Testing 1, 2, 3", "Testing 2, 3, 5", "Testing 3, 4, 7", "Testing 4, 5, 9", "Testing 5, 6, 11", "Testing 6, 7, 13", "Testing 7, 8, 15" ...
INPUT: "x=[INC_FLOAT], y=[INC_FLOAT=0.33], z=[INC_FLOAT=0.2,0.004]"
OUTPUT: "x=0.1, y=0.33, z=0.200", "x=0.2, y=0.43, z=0.204", "x=0.3, y=0.53, z=0.208", "x=0.4, y=0.63, z=0.212", "x=0.5, y=0.73, z=0.216", "x=0.6, y=0.83, z=0.220" ...
INPUT: "Season [PERIODIC=1,5], Episode [INTERVAL=1,5]"
OUTPUT: "Season 1, Episode 1", "Season 1, Episode 2", "Season 1, Episode 3", "Season 1, Episode 4", "Season 1, Episode 5", "Season 2, Episode 1", "Season 2, Episode 2", "Season 2, Episode 3" ...
```
<br><br>
Note that there might be spaces inside the tokens, for example you might find tokens like ```[INC_FLOAT = 0.2 , 0.500]```. All the patterns given as input are valid, there is no need to check for errors. If no tokens are found in the pattern, the generator simply generates the string every time the next function is called. | algorithms | import re
from itertools import *
def string_generator(pattern):
format = re . sub(
'\[.*?\]', '{}', pattern . replace('{', '{{'). replace('}', '}}'))
generators = [generator(x) for x in re . findall(
'\[.*?\]', pattern . replace(' ', ''))]
yield from starmap(format . format, zip(* generators)) if generators else repeat(pattern)
def generator(s):
match re . split('=|,', s[1: - 1]):
case["INC_INT", * args]: yield from inc_int(* map(int, args))
case["INC_FLOAT", * args]: yield from inc_float(precision(args), * map(float, args))
case["INTERVAL", * args]: yield from interval(* map(int, args))
case["PERIODIC", * args]: yield from periodic(* map(int, args))
def precision(args):
return max((len(x . split('.')[1]) for x in args), default=1)
def inc_int(start=1, step=1):
yield from count(start, step)
def inc_float(precision, start=0.1, step=0.1):
yield from map(f'{{ :. { precision } f}} ' . format, count(start, step))
def interval(first=1, last=None):
yield from cycle(range(first, (last or first) + 1))
def periodic(start=1, n=1):
yield from (start + x / / n for x in count())
| String generation by pattern | 62b3356dacf409000f53cab7 | [
"Algorithms",
"Strings",
"Regular Expressions"
] | https://www.codewars.com/kata/62b3356dacf409000f53cab7 | 5 kyu |
[Do the simpler version first!](https://www.codewars.com/kata/618647c4d01859002768bc15)
In this generalized version of the riddle, each 'player' has a hat and must guess what color it is. The players are lined up such that they can see the hats of everyone in front of them, but not their own hat or those of the players behind them. Each player must guess the color of their own hat, taking turns from back to front. In order to win, only one guess may be wrong.
The team will be told which colors of hats will be present (but not how many of each color) and given time to develop a strategy before they get their hats.
Your task is to create the team's strategy by writing a function that returns the guess for a player based on the information they have.
### Inputs
- a list of all colors of hat (possibly) present.
```if:haskell
**Colors can be any Eq type; not just Strings.**
```
- a list of previous guesses, starting from the back (empty for the first player).
- a list of the hats the player can see, starting with the one directly in front of them (empty for the last player).
### Output
- the guess, which should be one of the elements from the list of colors passed in
All inputs in the test cases will follow the rules of the riddle. Your solution *should* work for any number of colors, but the test cases will only go up to 5 colors and 100 players. Tests assume the first person to guess is wrong, since it is impossible to know what color hat they have.
| games | def guess(colours, guesses, hats):
return colours[- sum(colours . index(c) for c in guesses + hats) % len(colours)]
| Extreme Hat Game | 62bf879e8e54a4004b8c3a92 | [
"Riddles"
] | https://www.codewars.com/kata/62bf879e8e54a4004b8c3a92 | 5 kyu |
A common way to teach children how to code is to let them program a robot with a very limited set of instructions, which are executed in an endless loop.
In our case, you have to help a little robot by finding the correct set of instructions, so it can cross a dangerous area with holes.
The robot starts in the upper left corner and has to leave the area either on the right or the down side.
The robot can only handle two types of instructions:
- move one tile to the right: ```R```
- move one tile down: ```D```
Its memory is limited, therefore the number of instructions you can provide is limited ```max_instructions```.
After the end of the instructions is reached, the program starts at the beginning.
The solution can be shorter than the maximum number of instructions.
The area you have to cross is provided as a new line separated string. ```F``` denotes a movable tile, ```H``` denotes a hole, which will kill your robot, if it moves onto it. The field is always quadratic.
Your solution should provide the program for the robot as string.
All test cases have at least one solution.
### Input
```field``` the field your robot has to pass.
```max_instructions``` the maximum number of instructions the robot can handle
### Output
String containing the instructions ```R``` and ```D```
A very similar challenge can be found at https://www.hacker.org/runaway/
There is also version with higher perfomance requirements of this kata available: [Program the robot (performance)](https://www.codewars.com/kata/62cb28bfd32bd6184fd500fe/) | algorithms | from itertools import cycle
def program(field, max_instructions):
dirs, fld = {'R': (0, 1), 'D': (1, 0)}, field . split('\n')
ln = len(fld)
seen = {(0, 0)}
queue = [[(0, 0, '')]]
def is_path(path):
x, y = 0, 0
for p in cycle(path):
x, y = x + dirs[p][0], y + dirs[p][1]
if x >= ln or y >= ln:
return True
if fld[x][y] == 'H':
return False
for _ in range(max_instructions):
cur = []
for q in queue:
for k, (dx, dy) in dirs . items():
x, y = q[- 1][0] + dx, q[- 1][1] + dy
path = '' . join(x for _, _, x in q) + k
if x >= ln or y >= ln or is_path(path):
return path
if (x, y) not in seen and fld[x][y] != 'H':
cur . append(q + [(x, y, k)])
queue = cur[:]
| Program the robot | 62b0c1d358e471005d28ca7e | [
"Algorithms",
"Graph Theory"
] | https://www.codewars.com/kata/62b0c1d358e471005d28ca7e | 6 kyu |
## Overview
The usual way of dividing a pizza is fair but really boring; you place `n` points evenly spaced around the perimeter of the pizza, then cut towards the centre point - you always obtain a total of `n` equally sized, equally shaped pieces.
Suppose for your next pizza you do the following: you place `n` points **any way you like** around the perimeter of the pizza, and you cut through **all possible straight lines** formed by the `n` points.
What is the maximum number of pieces that you can cut your pizza into?
Note: the pieces **don't** have to be of the same size or shape. Also some pieces may be completely "internal" and not touch the perimeter's crust (perfect for people on a low-carb diet).
---
## Inputs and Outputs
You will be given an `integer`, `n`, representing the number of points you can place on the perimeter of your pizza.
The value of `n` will always be `n >= 2`; when you have `0` or `1` point you cannot divide the original pizza into more than the 1 original "piece".
Input values up to `n = 1000` will be tested in random tests.
You must return an `integer`, representing the maximum number of pieces you can cut your pizza into using the algorithm described above.
---
## Code golf requirement
Depending on language, your kata solution code must be less than or equal to the following number of characters:
- Python: `<= 42` characters
- JavaScript: `<= 42` characters
---
## Example with (magnificent) illustration
Click the thumbnail to see full-sized illustration:
<a href="https://ibb.co/bvMVCbH"><img src="https://i.ibb.co/bvMVCbH/pizza-kata-codewars.png" alt="pizza-kata-codewars" border="0"></a>
In this example with `n = 4` points placed around the perimeter of the pizza, you can create a maximum total of `8` pieces by cutting along all the `6` lines formed by the `n = 4` points.
Therefore, `pizza(4) = 8`. | reference | def pizza(n): return n * ~ - n * (n * n - 5 * n + 18) / / 24 + 1
| [Code golf] Strange way to divide a pizza | 62bd9616aced6376cc37950e | [
"Mathematics",
"Restricted"
] | https://www.codewars.com/kata/62bd9616aced6376cc37950e | 6 kyu |
### **The Cipher**
1. Take every letter in the one-word message and replace it with its place in the alphabet.:
2. For each number in the list, multiply it by 3 and subtract 5. Repeat this process *n* times.
## Task
Implement two functions, `encrypt()` and `decrypt()`.
#### `encrypt()` (details):
* Input:
* word - a word
* n - the key
* Output:
* *word* encrypted with ***The Cipher***
* Examples:
* `n = 2: "abc" >> [1, 2, 3] >> [-2, 1, 4] >> [-11, -2, 7]`
* `n = 6: "test" >> [12760, 1825, 12031, 12760]`
* `n = 3: "hi" >> [151, 178]`
#### `decrypt()` (details):
* Input:
* encrypted_word - A word that got encrypted with ***The Cipher***
* n - the key
* Output:
* *encrypted_word*, decrypted
* Examples:
* `n = 2: [16, 25, 34] >> [7, 10, 13] >> "def"`
* `n = 5: [1339, 610, 2311, 2311, 3040] >> "hello"`
### Notes
All the inputs in the test cases are valid, i. e. for `encrypt()` only lowercase letters and for `decrypt()` only lists that were actually encrypted by my own encrypt() function.
If *n* is negative, just treat it as if it were 0.
Credits to 2021 Gauss Grade 7 | reference | BASE = ord("a") - 1
def encrypt(word, n):
xs = [x - BASE for x in map(ord, word)]
for _ in range(n):
xs[:] = (x * 3 - 5 for x in xs)
return xs
def decrypt(encrypted_word, n):
xs = encrypted_word[:]
for _ in range(n):
xs[:] = ((x + 5) / / 3 for x in xs)
return "" . join(chr(x + BASE) for x in xs)
| Cipher from math problem | 62bdd252d8ba0e0057da326c | [
"Cryptography",
"Mathematics",
"Ciphers",
"Algorithms"
] | https://www.codewars.com/kata/62bdd252d8ba0e0057da326c | 7 kyu |
> __Randonneuring__ (also known as Audax in the UK, Australia and Brazil) is a long-distance cycling sport with its origins in audax cycling. In randonneuring, riders attempt courses of 200 km or more, passing through predetermined "__controls__" (checkpoints) every few tens of kilometres. Riders aim to complete the course within specified time limits, and receive equal recognition regardless of their finishing order.
--Wikipedia
Each __control__ is open during a limited time. This time is defined by start time and control location. For example, riders starting at `07:00` must appear at `250` km control between `14:27` and `23:40` on the same date.
Your task is to make two functions calculating control open and control close times.
__Control open__
time is the time after which the fastest riders are expected to arrive at __control__. It's calculated through the maximum expected average speed:
- `34` km/h over the initial `200` km
- `32` km/h over km `200` - `400`
- `30` km/h over km `400` - `600`
- `28` km/h over km `600` - `1000`
- `26` km/h over km `1000` - `1200`
- `25` km/h over km `1200` - `1800`
- `24` km/h over km `1800` - `2000`.
For example, riders starting at `07:00` are expected to appear at `250` km control after `14:27`. Let's see how we got that number:
- take start time `07:00`
- add `05:52:56.471` (`200`km divided by `34`km/h)
- add `01:33:45.000` (`50`km divided by `32`km/h)
- round `14:26:41.471` to `14:27`
__Control close__
time is the time by which the slowest riders have to arrive at __control__ or be disqualified. It's calculated through minimum required average speed:
- `20` km/h over the initial `60` km
- `15` km/h over km `60` - `600`
- `11.428` km/h over km `600` - `1000`
- `13.333` km/h over km `1000` - `1200`
- `11` km/h over km `1200` - `1400`
- `10` km/h over km `1400` - `1800`
- `9` km/h over km `1800` - `2000`.
- Since it can take a lot of time for big groups of riders to start, `1` extra hour is added to the result regardless of control location.
For example, riders starting at `07:00` are expected to appear at `250` km control before `23:40`. Let's see how we got that number:
- take start time `07:00`
- add `03:00:00.000` (`60`km divided by `20`km/h)
- add `12:40:00.000` (`190`km divided by `15`km/h)
- add `01:00:00.000` (an extra hour to compensate for the slow start)
- round `23:40:00.000` to `23:40`
Input:
- `startTime` (`start_time`) - a string representing start time in 24-hour format,`%H:%M`
- `km` - a whole number representing control location in km, `0` <= `km` <= `2000`
Output:
- a string representing control open(close) time in 24-hour format,`%H:%M`, rounded to a minute, with zero padding.
- if control open(close) time is not on the start date, reflect it after result in brackets, e.g. `07:00 (+1)` means '07:00 on the next day'. | reference | from datetime import datetime, timedelta
# Encode the given velocity and interval information
# (Velocity, start kms, stop kms)
FAST = [
(34, 0, 200),
(32, 200, 400),
(30, 400, 600),
(28, 600, 1000),
(26, 1000, 1200),
(25, 1200, 1800),
(24, 1800, 2000),
]
SLOW = [
(20, 0, 60),
(15, 60, 600),
(11.428, 600, 1000),
(13.333, 1000, 1200),
(11, 1200, 1400),
(10, 1400, 1800),
(9, 1800, 2000),
]
def duration_to_hms(duration):
"""Convert duration in decimal hours to hours, minutes, seconds to nearest second"""
hours = int(duration)
residue = duration - hours
mins = int(residue * 60)
residue -= mins / 60
secs = round(residue * 3600)
return hours, mins, secs
def calc_interval_and_durations(vel, start, stop):
"""Given velocity and start/stop distances calculate:
1. The interval.
2. The duration for full interval in decimal time.
3. The hours/minutes/seconds of the duration (to nearest second)
"""
interval = stop - start
duration = interval / vel
hours, mins, secs = duration_to_hms(duration)
return interval, duration, hours, mins, secs
# Convert values to something a little more useful for calculation later
fast = [
calc_interval_and_durations(vel, start, stop)
for vel, start, stop in FAST
]
slow = [
calc_interval_and_durations(vel, start, stop)
for vel, start, stop in SLOW
]
def hh_mm_to_time(s):
"""
Convert a string in the format HH:MM to hours and minutes after 2000-01-01 00:00:00.0
"""
hh, mm = s . split(':')
hh, mm = int(hh), int(mm)
return datetime(2000, 1, 1, hh, mm, 0)
def time_to_control(dt):
"""
Convert a timestamp into the control format
"""
if dt . second >= 30:
# Round up a minute
dt += timedelta(minutes=1)
hh_mm = f' { dt : % H : % M } '
if dt . day > 1:
return f' { hh_mm } (+ { dt . day - 1 } )'
return hh_mm
def calculate_control(stages, time, km):
"""
Given the stages, the start time and the distance calculate the minimum timestamp for alloted arrival
"""
timestamp = hh_mm_to_time(time)
for interval, duration, hours, mins, secs in stages:
if interval <= km:
timestamp += timedelta(hours=hours, minutes=mins, seconds=secs)
km -= interval
else:
hours, mins, secs = duration_to_hms(duration / interval * km)
timestamp += timedelta(hours=hours, minutes=mins, seconds=secs)
km = 0
if km == 0:
break
return timestamp
def control_open_time(time, km):
"""
Given start time and distance find the open control time to nearest minute
"""
timestamp = calculate_control(fast, time, km)
return time_to_control(timestamp)
def control_close_time(time, km):
"""
Given start time and distance find the close control time to nearest minute
"""
timestamp = calculate_control(slow, time, km)
# Allow for start time spread
timestamp += timedelta(hours=1)
return time_to_control(timestamp)
| Randonneuring control calculator | 62b708450ee74b00589fcaba | [
"Mathematics",
"Date Time",
"Algorithms"
] | https://www.codewars.com/kata/62b708450ee74b00589fcaba | 5 kyu |
<h2> Task </h2>
Let
`$gcd(a, b) = x \\ lcm(a, b) = y \\ a, b \in \N$`
where gcd represents the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of two integers and lcm represents the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of two integers.
Given `$x$` and `$y$`, find any possible pair `$(a, b)$` that satisfies the condition above with `$a$` and `$b$` being positive integers. **If there are multiple solutions, return any.** If there are no solutions, `return None`. Return your answer as a tuple of `$(a, b)$`
<h2> Constraints </h2>
`$ 1 \leq x, y \leq 10^9 $`
<h2> Examples (More in Example Tests) </h2>
```
---------------------------------
| x | y | a | b |
| 6 | 36 | 12 | 18 |
| 7 | 8 | None |
| 8 | 48 | 24 | 16 |
---------------------------------
```
Note that there are **multiple solutions** for the 1st and 3rd examples.
**If you are having issues about the time limit, please try to first make an observation instead of straight up brute force. View solutions if necessary (they may surprise you!)** | reference | # I mean, they work
def gcd_lcm(x, y):
if y % x == 0:
return x, y
| GCD and LCM | 62b931bcb16c630025076970 | [
"Performance",
"Algorithms",
"Mathematics"
] | https://www.codewars.com/kata/62b931bcb16c630025076970 | 7 kyu |
This kata is a harder version of this kata: https://www.codewars.com/kata/5727bb0fe81185ae62000ae3
If you haven't done it yet, you should do that one first before doing this one.
You are given a ```string``` of letters that you need to type out. In the string there is a special function: ```[backspace]```. Once you encounter a ```[backspace]``` , you delete the character right before it. If there is nothing to ```backspace``` , just carry on. Return the final ```string``` .
e.g. ```oopp[backspace]s``` return ```oops``` (delete the p)
Walkthrough:
o\
oo\
oop\
oopp\
oop ```[backspace]```\
```oops```
e.g. ```ooppp[backspace][backspace]s``` return ```oops``` (delete both p's)
Walkthrough:
o\
oo\
oop\
oopp\
ooppp\
oopp ```[backspace]```\
oop ```[backspace]```\
```oops```
e.g. ```a[backspace][backspace]ooppp[backspace][backspace]s``` return ```oops```
Walkthrough:
a\
[nothing]\
[nothing]\
o\
oo\
oop\
oopp\
ooppp\
oopp ```[backspace]```\
oop ```[backspace]```\
```oops```
But there's a twist! ```[backspace][backspace][backspace]``` can appear as ```[backspace]*3``` or even ```[backspace]*2[backspace]```
Basically, ```[backspace][backspace] ... [backspace]``` ```n times``` can appear as ```[backspace]*n``` (n can even be 1, but not less than 1 (`[backspace]*0` will not occur))
e.g. ```a[backspace]*2oopppp[backspace]*2[backspace]s``` return ```oops```
Walkthrough:
a\
[nothing] ```[backspace]*2```\
o\
oo\
oop\
oopp\
ooppp\
oopppp\
oopp ```[backspace]*2```\
oop ```[backspace]```\
```oops```
To make it easier, only letters, spaces. and the ```[backspace]``` function will be in the initial string.
Good luck!
______
Check out <a href="https://www.codewars.com/collections/65abaddba4b9f2013327dd3d">the rest of the kata in this series</a>! | algorithms | import re
def solve(s):
stk = []
for m in re . finditer(r'\[backspace\](?:\*(\d*))?|.', s):
if m[0]. startswith('[backspace]'):
n = int(m[1] or 1)
while n and stk:
stk . pop()
n -= 1
else:
stk . append(m[0])
return '' . join(stk)
| Typing series #1 --> the [backspace] function | 62b3e38df90eac002c7a983f | [
"Algorithms",
"Strings"
] | https://www.codewars.com/kata/62b3e38df90eac002c7a983f | 6 kyu |
Farmer Thomas is an edgy teenager who likes to ruin other people's dream of becoming a software engineer. Motivated by this, he attacked the Codewars servers with a strawberry baguette at 8:00 UTC on June 21 2022, and took down the Codewars website for an hour.
However, there was barely anyone online at that time, so Farmer Thomas' attack was very ineffective, as it affected very few people. Now, he wants your help in planning the next attack!
In the next `n` hours, the array `user_count` of length `n` stores the amount of people online at that hour. Farmer Thomas has enough money to attack the Codewars server for `k` hours (not necessarily in one consecutive attack though). He wants to plan the attack to affect as many people as possible.
However, the Codewars website has a backup battery that can keep the site running for two additional hours once an attack starts. This means that each period of attack must be more than 2 hours to have an actual effect. After the attack stops, Codewars will go back online and its backup battery will recharge instantly, ready for the next attack.
Take the following plan as example (`√` means Thomas is attacking at that hour, and `X` means not attacking):
```
k = 7
user_count: [ 30 10 0 100 200 300 20 10 30 1000 ]
should_attack: X X √ √ √ √ X √ √ √
affected_user_count: 200 + 300 + 1000 = 1500
```
In the above example, the sum of the duration of all the attacks is at most `k=7`, which can be arranged to be carried out anytime during the 10-hour time frame (as per the length of `user_count`). However, the first 2 hours in each consecutive period of attack do not affect the website (due to the backup battery) and thus do not add to the `total_affected_user`.
Given `user_count` and `k`, return the maximum amount of users that can be affected with the most optimal attacking plan. Good luck!
## Constraints
- `n < 500`
- `k < 250`
- 50 large tests
| algorithms | def attack_plan(user_count, k):
dp = [[[- float("inf")] * 4 for _ in range(k + 1)]
for _ in range(len(user_count) + 1)] # [time][attacks][consecutive]
dp[0][0][0] = 0
for i, count in enumerate(user_count):
for r in range(k + 1):
dp[i + 1][r][0] = max(dp[i + 1][r][0], max(dp[i][r])) # Stop attack
if r < k:
dp[i + 1][r + 1][1] = max(dp[i + 1][r + 1][1], dp[i][r][0])
dp[i + 1][r + 1][2] = max(dp[i + 1][r + 1][2], dp[i][r][1])
dp[i + 1][r + 1][3] = max(dp[i + 1][r + 1][3],
dp[i][r][2] + count, dp[i][r][3] + count)
return max(dp[- 1][- 1])
| Take Down Codewars! | 62b1860db38ba1739bcaaebf | [
"Dynamic Programming",
"Algorithms",
"Arrays"
] | https://www.codewars.com/kata/62b1860db38ba1739bcaaebf | 5 kyu |
## Story
YouTube had a like and a dislike button, which allowed users to express their opinions about particular content. It was set up in such a way that you cannot like and dislike a video at the same time.
There are two other interesting rules to be noted about the interface:
Pressing a button, which is already active, will undo your press.
If you press the like button after pressing the dislike button, the like button overwrites the previous "Dislike" state. The same is true for the other way round.
## Task
Create a function that takes in a list of button inputs and returns the final state.
## Examples
```python
like_or_dislike([Dislike]) ➞ Dislike
like_or_dislike([Like, Like]) ➞ Nothing
like_or_dislike([Dislike, Like]) ➞ Like
like_or_dislike([Like, Dislike, Dislike]) ➞ Nothing
```
```haskell
[Dislike] -> Just Dislike
[Like,Like] -> Nothing
[Dislike,Like] -> Just Like
[Like,Dislike,Dislike] -> Nothing
```
```factor
{ Dislike } -> Dislike
{ Like Like } -> Nothing
{ Dislike Like } -> Like
{ Like Dislike Dislike } -> Nothing
```
```javascript
likeOrDislike([Dislike]) => Dislike
likeOrDislike([Like,Like]) => Nothing
likeOrDislike([Dislike,Like]) => Like
likeOrDislike([Like,Dislike,Dislike]) => Nothing
```
```coffeescript
likeOrDislike([Dislike]) -> Dislike
likeOrDislike([Like,Like]) -> Nothing
likeOrDislike([Dislike,Like]) -> Like
likeOrDislike([Like,Dislike,Dislike]) -> Nothing
```
```lambdacalc
like-or-dislike [Dislike] -> Dislike
like-or-dislike [Like,Like] -> Nothing
like-or-dislike [Dislike,Like] -> Like
like-or-dislike [Like,Dislike,Dislike] -> Nothing
```
```prolog
like_or_dislike([dislike], dislike).
like_or_dislike([like, like], nothing).
like_or_dislike([dislike, like], like).
like_or_dislike([like, dislike, dislike], nothing).
```
```julia
likeordislike([Dislike]) => Dislike
likeordislike([Like, Like]) => nothing
likeordislike([Dislike, Like]) => Like
likeordislike([Like, Dislike, Dislike]) => nothing
```
## Notes
- If no button is currently active, return `Nothing`.
- If the list is empty, return `Nothing`.
~~~if:lambdacalc
## Encodings
purity: `LetRec`
numEncoding: `None`
export constructors `nil, cons` for your `List` encoding
you can use `Like, Dislike, Nothing` which are `Preloaded` as
```lambdacalc
Like = λ like dislike nothing . like
Dislike = λ like dislike nothing . dislike
Nothing = λ like dislike nothing . nothing
```
~~~
| games | def like_or_dislike(lst):
state = 'Nothing'
for i in lst :
state = 'Nothing' if i == state else i
return state | Likes Vs Dislikes | 62ad72443809a4006998218a | [
"Arrays",
"Algorithms"
] | https://www.codewars.com/kata/62ad72443809a4006998218a | 7 kyu |
# Consecutive Vowels #
You are given a random string of lower-case letters. Your job is to find out how many ordered and consecutive vowels there are in the given string beginning from `'a'`. Keep in mind that the consecutive vowel to `'u'` is `'a'` and the cycle continues.
Return an integer with the length of the consecutive vowels found.
This is better explained with a couple of examples:
1. You are given the string `"agrtertyfikfmroyrntbvsukldkfa"`. The logic is that you start from the `'a'` and make your way right. The next vowel is an `'e'` and it is the consecutive vowel, then `'i'` and so forth until you get to `'u'`. If you keep moving right you find `'a'` which happens to be the consecutive vowel. Return `6`.
1. This is a slightly trickier example: you are given the string `"erfaiekjudhyfimngukduo"`. As you move right you ignore all vowels until you get to the `'a'`, then ignore the rest until you get to the `'e'`, which is the consecutive vowel, and so forth until you get to the `'o'`. Return `4`.
# Note #
For this kata, the vowels are `'a', 'e', 'i', 'o', 'u'`, in that order. `'y'` is not considered a vowel in this kata.
Good luck! | reference | def get_the_vowels(word):
n = 0
for i in word:
if i == "aeiou" [n % 5]:
n += 1
return n
| Consecutive Vowels in a String | 62a933d6d6deb7001093de16 | [
"Strings",
"Fundamentals"
] | https://www.codewars.com/kata/62a933d6d6deb7001093de16 | 7 kyu |
Sergeant Ryan is a soldier in the orange army, fighting a war against the green army. One day, while out on patrol, Ryan notices [smoke signals](https://en.wikipedia.org/wiki/Smoke_signal) being used in the forest, presumably by green soldiers. Ryan never enjoyed solving cryptography problems in school, so that's why he needs your help to decode the signals.
The smoke signals come in the form of puffs of smoke, followed by a gap of clear air. There may be multiple sequences of puffs before the message ends. Ryan has concluded that the time between puffs of smoke is not important, so he writes down the sequences of puffs, separated by dots, such as:
```
5 (5 puffs of smoke)
3.4 (3 puffs followed by 4 puffs)
2.1.2.4 (2 puffs, 1 puff, 2 puffs, 4 puffs)
```
All of the smoke signals are transmitted in the morning, and they correspond to events that happen later in the day. Ryan believes that each signal directly corresponds to an event that occurs later that day. For example:
```
signals = ["2.1","3"]
events = ["Ambush in the jungle","Airstrike on camp golf"]
```
*(Ryan knows that the signals 2.1 and 3 must correspond to the above events, but he doesn't know which one belongs to which - he'll need more information from another day's signals and events)*
**Help Sergeant Ryan develop an algorithm to automatically decode the enemy smoke signals and return a dictionary that relates signals to events**, for example:
```
decoded = {
"2.1": "Airstrike on camp golf",
"3": "Ambush in the jungle"
}
```
Clarifications / Hints:
- You will receive one or more day's worth of signals and events to process
- Each smoke signal will correspond to exactly one event
- Multiple types of signals may correspond to the same type of event
- The ordering of events in a day does not necessarily follow the ordering of signals - if they did, Ryan would be able to solve this task without you!
- All input will be correct and reasonable - you do not have to worry about edge cases
- There will be no ambiguity about the results - you will always have enough information to perfectly decode the smoke signals
- There are 60 random tests that consist of 4-12 unique signal types and up to 20 days worth of information. Your solution should be efficient enough to handle these tests. | algorithms | from copy import deepcopy
def decode_smoke_signals(days):
days = deepcopy(days)
decoded = {}
# Keep eliminating signals until the puzzle has been solved
changed = True
while changed:
# Go through day by day and attempt to determine what each signal could be.
changed = False
signals = {}
for day in days:
for signal in day[0]:
if not signal in signals:
# Add all of today's events as possibilities for this signal
signals[signal] = set([a for a in day[1]])
else:
# Remove any previous possibilities that aren't in today's events
signals[signal] = signals[signal]. intersection(set(day[1]))
# Add any signals that were solved to the decoded list and remove them
# from the days list (and up to 1 instance of each corresponding event)
for signal, possible in signals . items():
if len(possible) == 1:
changed = True
event = list(possible)[0]
decoded[signal] = event
for day in days:
if signal in day[0]:
day[0]. remove(signal)
day[1]. remove(event)
return decoded
| Smoke Signals | 62a3855fcaec090025ed2a9a | [
"Ciphers",
"Cryptography",
"Logic"
] | https://www.codewars.com/kata/62a3855fcaec090025ed2a9a | 5 kyu |
Given an integer `n` and two other values, build an array of size `n` filled with these two values alternating.
## Examples
```ruby
5, true, false --> [true, false, true, false, true]
10, "blue", "red" --> ["blue", "red", "blue", "red", "blue", "red", "blue", "red", "blue", "red"]
0, "one", "two" --> []
```
Good luck! | reference | def alternate(n, first_value, second_value):
return [[first_value, second_value][i % 2] for i in range(n)]
| Length and two values. | 62a611067274990047f431a8 | [
"Fundamentals",
"Arrays"
] | https://www.codewars.com/kata/62a611067274990047f431a8 | 7 kyu |
We want to find the area of a set of buildings stacked close to each other.
A matrix (an array of arrays) gives us a description of the building from above, where each number represents the height expressed in cubes `1*1*1`, the cubes are stacked adjacent to each other.
For example:
```
matrix = [[2,1],[2,0]]
In position (0,0), there are 2 cubes stacked on each other
In position (0,1), there is only 1 cube
In position (1,0), there are 2 cubes stacked on each other
In position (1,1), there are 0 cubes stacked on each other.
So, the total covered area is 20.
```
The matrix will be of at least size `1*1` and square.
**Note: The floor of the building counts as area as well.** | games | import numpy as np
def surface_area(m):
return 2 * np . count_nonzero(m) + sum(abs(np . diff(m, 1, i, 0, 0)). sum() for i in range(2))
| Skyscrapers Area | 62a25309b070a3002f53b684 | [
"Algorithms",
"Arrays"
] | https://www.codewars.com/kata/62a25309b070a3002f53b684 | 6 kyu |
### Description
It is well known that
`$\displaystyle\sum^n_{i=1}i = \frac{1}{2} n^2 + \frac{1}{2} n$`
Which can be rewritten as
`$\displaystyle 1+2+3+4+...n = \frac{1}{2} n^2 + \frac{1}{2} n$`
Also
`$\displaystyle\sum^n_{i=1}i^2 = \frac{1}{3} n^3 + \frac{1}{2} n^2 + \frac{1}{6} n$`
Which can be rewritten as
`$\displaystyle 1^2+2^2+3^2+4^2+...n^2 = \frac{1}{3} n^3 + \frac{1}{2} n^2 + \frac{1}{6}n$`
### Task:
`$\displaystyle\sum^n_{i=1}i^k$`
Given a natural number value for `$k$` for the equation above, generate a simplified expanded form equation that would be equivalent to the summation.
Make sure your answer has terms with higher degree coming first.
For a better idea of the exact format, refer to the examples.
#### Notice
```if:python
Test cases will get large as `$k$` can go up to 140
```
```if:java
Test cases will get large as `$k$` can go up to 140, so you might consider using BigInteger
```
```if:javascript
Test cases will get large as `$k$` can go up to 140, so you might consider using BigInteger
```
#### Example 1
`equation(1) -> "1/2n^2 + 1/2n"`
#### Explanation
`$\displaystyle\sum^n_{i=1}i^1 = \frac{1}{2} n^2 + \frac{1}{2} n$`
#### Example 2
`equation(4) -> "1/5n^5 + 1/2n^4 + 1/3n^3 - 1/30n"`
#### Explanation
`$\displaystyle\sum^n_{i=1}i^4 = \frac{1}{5} n^5 + \frac{1}{2} n^4 + \frac{1}{3} n^3 - \frac{1}{30} n$` | reference | from fractions import Fraction
from functools import lru_cache
from math import comb, factorial
B = [Fraction(1, 1)]
for i in range(1, 141):
B . append(
1 - sum(comb(i, j) * Fraction(B[j], i - j + 1) for j in range(i)))
def build(
i, x, y): return f" { abs ( x )}{ f'/ { y } ' * ( y > 1 )} n { f'^ { i } ' * ( i > 1 )} "
fact = lru_cache(maxsize=None)(factorial)
def equation(k):
coeff = [build(k + 1, 1, k + 1), '+', build(k, 1, 2)]
for i in range(2, k + 1):
if (x := (B[i] * fact(k)) / (fact(i) * fact(k - i + 1))):
coeff . append("-+" [x > 0])
coeff . append(build(k - i + 1, * x . as_integer_ratio()))
return ' ' . join(coeff)
| n^k summation | 62a0a24ed518853c3528683f | [
"Mathematics"
] | https://www.codewars.com/kata/62a0a24ed518853c3528683f | 4 kyu |
## Overview
We are going to be studying permutations of `n` distinct elements - for simplicity we shall use the integers from `1` to `n`.
If we start with the integers `1` to `n` in ascending order in a list:
`[1, 2, 3, ..., n-2, n-1, n]`
then, for a given value of `n`, how many permutations of this list are there in which each element's index changes by **AT MOST** 1 after the permutation?
---
## Inputs
You will be given an `integer`, `n`, representing the number of distinct elements that we want to permute.
As mentioned, for definiteness, you can think of the integers from `1` to `n` in a starting list given by `[1, 2, ..., n]`
**Performance requirement:** Values up to `n = 200` will be tested.
## Output
You must return an `integer` representing the total number of possible permutations of the list `[1, 2, ..., n]` that satisfy the requirement.
## Important notes
1. The identity permutation, where no element changes its index, is a valid permutation according to the definition, so make sure you include it in your total.
2. You **cannot** wrap-around the list, so changing element `1` with element `n` is **not allowed** (except for the case `n = 2` of course).
---
## Walk-through Example
Let's consider the case `n = 4`; so concretely we are looking at permutations of the list `[1,2,3,4]`
There are in total `n! = 4! = 24` possible permutations of this list.
If you write out all of these 24 possible permutations, you will find that only `5` satisfy the above requirement:
`[1,2,3,4]` - no element changes index.
`[2,1,3,4]` - elements 1&2 change indices by +1 and -1 respectively.
`[1,3,2,4]` - elements 2&3 change indices by +1 and -1 respectively.
`[1,2,4,3]` - elements 3&4 change indices by +1 and -1 respectively.
`[2,1,4,3]` - elements 1&2 change indices by +1 and -1 respectively **and** elements 3&4 change indices by +1 and -1 respectively.
In all of the `5` above permutations, the elements' indices have indeed changed by at most 1.
So for the input `n = 4` you should return the answer `5`.
**Explanation of an INVALID permutation for n = 4**
To be clear, let's consider another permutation which **does not** satisfy the kata's requirement, and therefore should not be included in your total:
`[3,4,2,1]` - element 1 has changed index by +3, element 2 by +1, element 3 by -2, element 4 by -2.
The above permutation is therefore **not** allowed, since at least one element has changed index by **more than** 1. | reference | def permuts(n):
x = [1, 2]
for i in range(n):
x . append(x[- 1] + x[- 2])
return x[n - 1]
| How many permutations where indices change by at most 1 | 629e18298f2d21006516e381 | [
"Fundamentals",
"Algorithms",
"Puzzles",
"Mathematics",
"Discrete Mathematics"
] | https://www.codewars.com/kata/629e18298f2d21006516e381 | 6 kyu |
**Description**
A random place has a currency system that consists of 2 coins
You are given the values of the coins, they can be identical, and they will always be positive.
Find the largest value that any whole number combination of the two
cannot produce
return `-1` if there is no such value or if it is an infinite number
**Example 1**
`coin1 = 1, coin2 = 1` -> `-1`
***Explanation***
You can produce any whole number with this combination
**Example 2**
`coin1 = 2, coin2 = 3` -> `1`
***Explanation***
no whole number combination of a
2 cent and 3 cent coin can produce 1 cent,
but they can produce every other number above 1
**Example 3**
`coin1 = 2, coin2 = 2` -> `-1`
***Explanation***
This coin combination will never produce an odd number, henceforth
the largest coin value this combination can not produce is some infinite odd number
therefore you return `-1` | reference | from math import gcd
def coins(coin1, coin2):
return - 1 if gcd(coin1, coin2) != 1 else (coin1 - 1) * (coin2 - 1) - 1
| Coins | 620dd259f7b0000017fc7b45 | [
"Mathematics",
"Algorithms"
] | https://www.codewars.com/kata/620dd259f7b0000017fc7b45 | 6 kyu |
### Introduction
**The Robber Language** (Rövarspråket) is a Swedish *cant* used by children to mislead people not familiar with the language. It's used by kids in several screenplays created by [Astrid Lindgren](https://en.wikipedia.org/wiki/Astrid_Lindgren).
The idea is that every consonant in a sentence is **duplicated**, with an `o` **inserted** in between (`b` becomes `bob`). Vowels are left untouched (`a` remains an `a`). It's quite hard for an untrained ear to understand a conversation encoded in this manner, especially if spoken rapidly.
### Example
In this example, the **consonants** `f`, `b` and `r` are modified. The **vowels** `o` and `a` are left untouched.
```
'foo bar' => 'fofoo bobaror' (`fof-o-o bob-a-ror`)
```
**Note**: The dashes `-` in the example are added for readability and should **not** be included in the output.
### The Assignment
~~~if:c
Your task is to implement a function `char* robber_encode(const char *sentence, char encoded[256])` which takes in a string `sentence` and writes the result, converted into **robber language**, in the output parameter `encoded` (a pre-allocated character buffer large enough to hold the result). The function returns a pointer to the string `encoded`.
~~~
~~~if:python
Your task is to implement a function `def robber_encode(sentence)` which takes in a string `sentence` and returns the result, converted into **robber language**, as a string.
~~~
~~~if:javascript
Your task is to implement a function `function robberEncode(sentence)` which takes in a string `sentence` and returns the result, converted into **robber language**, as a string.
~~~
~~~if:csharp
Your task is to implement a function `RobberEncode(string sentence)` which takes in a string `sentence` and returns the result, converted into **robber language**, as a string.
~~~
### Notes
1. Only **consonants** should be modified; leave all other characters unchanged.
2. Both **upper** and **lower** case characters will be tested.
3. The **case** of the `o` character and the neighboring consonants should match (`F => FOF` and `f => fof`).
4. For the purpose of this kata, a character is considered a **consonant** if it's equal to one of the letters `BCDFGHJKLMNPQRSTVWXYZ`.
### Testing
The tests will challenge your function with sentences of **length 0-255**, ASCII characters between **32-126** and ***nothing else***.
### Links
Check out these links for more information:
* https://en.wikipedia.org/wiki/R%C3%B6varspr%C3%A5ket
| reference | from functools import partial
from re import compile
r1 = partial(compile(r"(?=[a-z])([^aeiou])"). sub, r"\1o\1")
r2 = partial(compile(r"(?=[A-Z])([^AEIOU])"). sub, r"\1O\1")
def robber_encode(sentence):
return r1(r2(sentence))
| The Robber Language | 629e4d5f24b98110a83b2d0d | [
"Fundamentals",
"Strings",
"Algorithms"
] | https://www.codewars.com/kata/629e4d5f24b98110a83b2d0d | 7 kyu |
# Task
Create a decorator `@predicate` which allows boolean functions to be conveniently combined using `&`, `|` and `~` operators:
```python
@predicate
def is_even(num):
return num % 2 == 0
@predicate
def is_positive(num):
return num > 0
(is_even & is_positive)(4) # True
(is_even & is_positive)(3) # False
(is_even | is_positive)(3) # True
(~is_even & is_positive)(3) # True
```
It should work with all functions, regardless of how many arguments they accept:
```python
@predicate
def to_be():
return True
(to_be | ~to_be)() # True
@predicate
def is_equal(a, b):
return a == b
@predicate
def is_less_than(a, b):
return a < b
(is_less_than | is_equal)(1, 2) # True
```
Keyword arguments should work as well:
```python
(is_less_than | is_equal)(2, b=2) # True
(is_less_than | is_equal)(a=3, b=2) # False
```
Combinations of functions with incompatible signatures (e.g. `is_positive & is_less_than`) will not be tested.
A decorated function should be callable by itself (without combining with other predicates) and behave like the original function:
```python
@predicate
def is_less_than(a, b):
return a < b
is_less_than(1, 2) # True
is_less_than(2, 2) # False
is_less_than(3, 2) # False
```
Good luck!
This kata is heavily inspired by [FArekkusu](https://www.codewars.com/users/FArekkusu)'s [Readable Specification Pattern](https://www.codewars.com/kata/5dc424122c135e001499d0e5), but should be a bit easier. You should try his kata as well! | reference | class predicate:
def __init__(self, func): self . f = func
def __call__(self, * a, * * kw): return self . f(* a, * * kw)
def __invert__(self): return self . __class__(lambda * a, * * kw: not self . f(* a, * * kw))
def __or__(self, o): return self . __class__(lambda * a, * * kw: self . f(* a, * * kw) or o . f(* a, * * kw))
def __and__(self, o): return self . __class__(lambda * a, * * kw: self . f(* a, * * kw) and o . f(* a, * * kw))
| Combining predicates | 626a887e8a33feabd6ad8f25 | [
"Decorator",
"Functional Programming",
"Object-oriented Programming"
] | https://www.codewars.com/kata/626a887e8a33feabd6ad8f25 | 5 kyu |
This Kata will consider the situation of a fictional politician, "The Big Guy", or TBG.
TBG's close relative, "The Bagman", travels around the world collecting bribes for TBG.
Because of increased scrutiny, they can no longer use government aircraft to collect the bribes so they
have to pay for private travel. Any travel costs come out of the total rake, so TBG wants to keep travel costs at a minimum. <br>
<p> Neither TBG nor The Bagman is very good at math, so TBG has hired you to write an algorithm to determine the cheapest route for The Bagman
to take so that he can collect all the bribes. The route will always start and end in TBG's home city, Notgnihsaw.
Your function should return a list of the optimal path, in order, starting with the first city The Bagman travels to and ending with Notgnihsaw.
You will be provided two lists as inputs. The first will be the list of cities that The Bagman must travel to. The second will be a
2D list consisting of travel costs from the corresponding city in List 1 to the various other cities in the order they are given.
For example, given: <br>
cities = ['Notgnihsaw', 'Berlin', 'Helsinki'] <br>
costs = [[0, 800, 1500], [900, 0, 350], [1200, 650, 0]] <br>
From Notgnihsaw to Notgnihsaw costs 0, Notgnihsaw to Berlin costs 800, Notgnihsaw to Helsinki costs 1500.
From Berlin to Notgnihsaw costs 900, Berlin to Berlin costs 0, Berlin to Helsinki costs 350, etc.
Note that the cost of travelling from A to B is not necessarily the same as travelling B to A. <br>
Business is good, so there will always be at least 3 cities in the list (including Notgnihsaw). <br>
If there are multiple routes tied for the cheapest, any of them can be returned as the correct answer.
Example: <br>
cities = ['Notgnihsaw', 'Berlin', 'Helsinki'] <br>
costs = [[0, 800, 1500], [900, 0, 350], [1200, 650, 0]] <br>
returns ['Berlin', 'Helsinki', 'Notgnihsaw']
cities = ['Aleppo', 'Shenyang', 'Notgnihsaw', 'Vienna', 'Buenos Aires'] <br>
costs = [[0, 1800, 1250, 1500, 2450], [1400, 0, 1900, 1150, 2000], [1300, 1200, 0, 900, 1450], <br>
[3000, 1950, 800, 0, 1700], [2800, 2400, 1650, 2250, 0]] <br>
returns ['Shenyang', 'Aleppo', 'Vienna', 'Buenos Aires', 'Notgnihsaw'] <br>
Good luck, The Big Guy is counting on you! | algorithms | from itertools import permutations
def best_route(cities, costs):
nothome = [x for x in cities if x != 'Notgnihsaw']
tbl = {city1: {city2: cost for city2, cost in zip(
cities, row)} for city1, row in zip(cities, costs)}
return min(([* p, 'Notgnihsaw'] for p in permutations(nothome)), key=lambda tour: sum(tbl[a][b] for a, b in zip(['Notgnihsaw'] + tour, tour)))
| The Big Guy and The Bagman | 6297d639de3969003e13e149 | [
"Permutations",
"Algorithms"
] | https://www.codewars.com/kata/6297d639de3969003e13e149 | 5 kyu |
### Introduction
In this kata we are going to take a look at how IPv4 packet filtering works using **A**ccess **C**ontrol **L**ists (ACL), specifically using a technique called **wildcard masks** to match based on an IP address or a range of addresses.
A wildcard mask is the **inverse** of the subnet mask:
```
255.255.255.0 = 11111111.11111111.11111111.00000000 (subnet mask)
0.0.0.255 = 00000000.00000000.00000000.11111111 (wildcard mask)
```
Using the wildcard mask we can decide which bits to look at in the IPv4 address when filtering packets. If a bit in the mask is:
* A zero (`0`), that same bit **must match exactly** between the *IPv4 address* **and** the *network address*.
* A one (`1`), that bit can be **either a one or a zero** in the *IPv4 address* compared to the *network address*.
### Example
Let's look at an example:
```
permit 192.168.0.0 0.0.0.255
```
This is an entry in an **A**ccess **C**ontrol **L**ist (ACL) telling us to permit packets from `192.168.0.0-255`.
```
192.168.0.0 = 11000000.10101000.00000000.00000000 (network address)
0.0.0.255 = 00000000.00000000.00000000.11111111 (wildcard mask)
```
Again, the zeros (0) must **match exactly** and the ones (1) can be **anything**, so:
```
192.168.0.0 = 11000000.10101000.00000000.00000000 (match)
192.168.0.1 = 11000000.10101000.00000000.00000001 (match)
192.168.0.254 = 11000000.10101000.00000000.11111110 (match)
192.168.0.255 = 11000000.10101000.00000000.11111111 (match)
192.168.1.1 = 11000000.10101000.00000001.00000001 (mismatch)
192.168.1.254 = 11000000.10101000.00000001.11111110 (mismatch)
128.168.0.1 = 10000000.10101000.00000000.00000001 (mismatch)
193.168.0.1 = 11000001.10101000.00000000.00000001 (mismatch)
```
**Hint:** Using **bitwise operators** might be helpful here.
### The Assignment
~~~if:c
Your task is to implement a function `bool match(const char *net_addr, const char *wc_mask, const char *ipv4_addr)` which takes in a network address `net_addr`, a wildcard mask `wc_mask`, an IPv4 address `ipv4_addr` and determines if the IPv4 address is a match or mismatch. The function returns `true` if it's a match and `false` if it's a mismatch.
~~~
~~~if:python
Your task is to implement a function `match(net_addr: str, wc_mask: str, ipv4_addr: str) -> bool` which takes in a network address `net_addr`, a wildcard mask `wc_mask`, an IPv4 address `ipv4_addr` and determines if the IPv4 address is a match or mismatch. The function returns `True` if it's a match and `False` if it's a mismatch.
~~~
### Testing
The tests will challenge your function with addresses and masks in the full 32-bit range (`0.0.0.0 - 255.255.255.255`) and ***nothing else***.
### Links
Check out these links for more information:
* https://en.wikipedia.org/wiki/IPv4#Addressing
* https://en.wikipedia.org/wiki/Binary_number
* https://community.cisco.com/t5/networking-documents/access-control-lists-acl-explained/ta-p/4182349
| reference | from ipaddress import IPv4Address as ip
def match(net_addr: str, wc_mask: str, ipv4_addr: str) - > bool:
return int(ip(net_addr)) | int(ip(wc_mask)) == int(ip(ipv4_addr)) | int(ip(wc_mask))
| Networking 101/5: Access Control using a Wildcard mask | 6294a4d0eb816e36363b9079 | [
"Networks",
"Bits",
"Fundamentals",
"Binary"
] | https://www.codewars.com/kata/6294a4d0eb816e36363b9079 | 6 kyu |
### Introduction
An IPv4 address and its associated subnet mask is often written as
`192.168.0.7 255.255.255.0`. This combination tells us what the *address of the host* is (`192.168.0.7`) and its *network prefix*, in other words *which network* it belongs to (`192.168.0.0`).
Both the IPv4 address and subnet mask is made up of 32 bits. The bits are divided into four sections called octets (groups of eight). Each octet can have a number in the range 0 to 255 (unsigned 8-bit integers).
Let's take a closer look at the IPv4 information above.
```
192.168.0.7 = 11000000.10101000.00000000.00000111
255.255.255.0 = 11111111.11111111.11111111.00000000
|------------------------| |------|
network prefix hosts
```
Here we have the host address `192.168.0.7` which belongs to the network with a *network prefix* of `192.168.0.0/24` (CIDR notation).
In this kata we will focus on two things:
1. The number of *bits set to one* (1) in the netmask, which is **24** in our example above.
2. Which *network prefix* the netmask **masks out of the IPv4 address**, which is `192.168.0.0` (`11000000.10101000.00000000.00000000`) in this case.
If this seems a bit confusing it might get a bit clearer with another example, `192.168.0.21 255.255.255.248`:
```
11000000.10101000.00000000.00010101 (192.168.0.21)
& 11111111.11111111.11111111.11111000 (255.255.248.0)
= 11000000.10101000.00000000.00010000 (192.168.0.16)
|------------------------------||-|
network prefix hosts
```
Here we have the host address `192.168.0.21` which belongs to the network with a *network prefix* of `192.168.0.16/29` (CIDR notation).
**Hint**: Using *bitwise operators* might be helpful.
### CIDR Notation
**CIDR** is short for *Classless Inter-Domain Routing* and is, among other things, a way for us to write an IPv4 address together with its subnet mask in a more efficient way.
Using the first example above, we can write the address as:
```
192.168.0.7 255.255.255.0 (address in Classful notation)
192.168.0.7/24 (address in CIDR notation)
192.168.0.0/24 (prefix in CIDR notation)
```
We are simply showing the IPv4 address information together with the number of bits set in the netmask, which is **24** in this example.
### The Assignment
~~~if:c
Your task is to implement a function `void network_cidr(const char *ipv4_addr, const char *netmask, char result[19])` which takes in an IPv4 address `ipv4_addr` and a subnet mask `netmask`, and writes the result (the ***network prefix in CIDR notation***) into the output parameter `result`, a pre-allocated character buffer large enough to hold the result. The function has no return value.
~~~
~~~if:python
Your task is to implement a function `def network_cidr(ipv4_addr: str, net_mask: str) -> str` which takes in an IPv4 address `ipv4_addr` and a subnet mask `net_mask`, and returns the result (the ***network prefix in CIDR notation***) as a string.
~~~
### Testing
The tests will challenge your function with IPv4 addresses and subnet masks in the full 32-bit range (`0.0.0.0 - 255.255.255.255`) and ***nothing else***.
### Links
Check out these links for more information:
* https://en.wikipedia.org/wiki/Classless_Inter-Domain_Routing#CIDR_notation
* https://en.wikipedia.org/wiki/IPv4#Addressing
* https://en.wikipedia.org/wiki/Binary_number | reference | from ipaddress import IPv4Network
def network_cidr(ipv4_addr, net_mask):
return str(IPv4Network(f' { ipv4_addr } / { net_mask } ', strict=False))
| Networking 101/4: Determine the IPv4 network prefix in CIDR notation | 628f8211840e153e14346eaa | [
"Networks",
"Bits",
"Fundamentals",
"Binary"
] | https://www.codewars.com/kata/628f8211840e153e14346eaa | 6 kyu |
# Task
I need a wristband. Help me in identifying an actual wristband.
A wristband can have 4 patterns:
**horizontal:** each item in a row is identical.
**vertical:** each item in each column is identical.
**diagonal left:** each item is identical to the one on it's upper left or bottom right.
**diagonal right:** each item is identical to the one on it's upper right or bottom left.
Write a function that returns **True** if the section can be correctly classified into one of the 4 types, and **False** otherwise.
### Examples
#### Part of horizontal wristband
```
is_wristband([
["A", "A"],
["B", "B"],
["C", "C"]
]) # True
```
#### Part of vertical wristband
```
is_wristband([
["A", "B"],
["A", "B"],
["A", "B"]
]) # True
```
#### Part of diagonal left wristband
```
is_wristband([
["A", "B", "C"],
["C", "A", "B"],
["B", "C", "A"],
["A", "B", "C"]
]) # True
```
#### Part of diagonal right wristband
```
is_wristband([
["A", "B", "C"],
["B", "C", "A"],
["C", "A", "B"],
["A", "B", "A"]
]) # True
```
### Notes
- All inputs will be valid
- 1<= len(arr) <=2000
| algorithms | def is_wristband(arr):
note = {1, 2, 3, 4}
for i, j in zip(arr, arr[1:]):
if 1 in note and (len(set(i)) != 1 or len(set(j)) != 1):
note . remove(1)
if 2 in note and i != j:
note . remove(2)
if 3 in note and i[: - 1] != j[1:]:
note . remove(3)
if 4 in note and j[: - 1] != i[1:]:
note . remove(4)
if not note:
return False
return True
| Wristband Patterns | 62949a3deb816e36363b8ee6 | [
"Arrays",
"Algorithms"
] | https://www.codewars.com/kata/62949a3deb816e36363b8ee6 | 5 kyu |
## Task
Write a function that checks if two non-negative integers make an "interlocking binary pair".
## Interlock ?
* numbers can be interlocked if their binary representations have no `1`'s in the same place
* comparisons are made by bit position, starting from right to left (see the examples below)
* when representations are of different lengths, the unmatched left-most bits are ignored
## Examples
1) `a = 9`, `b = 4`
Stacking representations shows how they can interlock.
```
9 1001
4 100
```
Here, no `1`'s share any position, so the function returns `true`.
2) `a = 3`, `b = 6`
These representations do not interlock.
```
3 11
6 110
```
Finding they both have a `1` in the same position, the function returns `false`.
## Input
Two non-negative integers.
## Output
Boolean `true` or `false` whether or not these integers are interlockable.
## Enjoy!
Consider one of the following kata to solve next:
* [Playing With Toy Blocks ~ Can you build a 4x4 square?](https://www.codewars.com/kata/5cab471da732b30018968071)
* [Four Letter Words ~ Mutations](https://www.codewars.com/kata/5cb5eb1f03c3ff4778402099)
* [Crossword Puzzle! (2x2)](https://www.codewars.com/kata/5c658c2dd1574532507da30b)
* [Is Sator Square?](https://www.codewars.com/kata/5cb7baa989b1c50014a53333)
# Nota Bene:
This kata is accepting of translations for any languages other than: Java, JavaScript, CoffeeScript, TypeScript, Go, Groovy, Julia, Dart, and Kotlin; as those are currently underway by the author. Thank you! | algorithms | def interlockable(a, b):
return not a & b
| Interlocking Binary Pairs | 628e3ee2e1daf90030239e8a | [
"Binary",
"Bits",
"Algorithms"
] | https://www.codewars.com/kata/628e3ee2e1daf90030239e8a | 7 kyu |
The target dose for the oral medication *Katamol* has been calculated: <code>d</code> (in mg).
The only options to prescribe the medication are pills of strength <code>a</code> (in mg) or <code>b</code> (in mg), so it may not be possible to prescribe the exact target dose.
For a given target dose <code>d</code>, return the highest possible dose that can be made with any amount of pills <code>a</code> and/or <code>b</code> that does not exceed the target dose.
Input: 3 integers, <code>d</code>,<code>a</code>,<code>b</code> representing the calculated target dose, the dose of pill <code>a</code>, and the dose of pill <code>b</code> (all in mg)
Output: Integer value of the closest dose to the target that can be made with any amount of pills <code>a</code> and/or <code>b</code> without going above the target dose.
Example:
```if:python
Input:
99,25,60
Output:
85
# 85 can be made with 1 * 25mg pill and 1* 60mg pill, no closer dose can be made that is less than or equal to 99
```
Constraints:
<code>a</code> <= <code>d</code>
2 <=<code>d</code> <=10000
1 <= <code>a</code> < <code>b</code> <5000
| reference | def prescribe(d, a, b):
return max(x + (d - x) / / a * a for x in range(0, d + 1, b))
| Pill Pusher | 628e6f112324192c65cd8c97 | [
"Fundamentals"
] | https://www.codewars.com/kata/628e6f112324192c65cd8c97 | 6 kyu |
<h2>Nonogram encoder</h2>
If you're not familiar with nonograms, I recommend you to check this wikipedia page: <a href="https://en.wikipedia.org/wiki/Nonogram">Nonogram - Wikipedia</a>
My friend Alex really likes to solve nonograms, and I've drawn one for him:
<img src="https://i.imgur.com/IwaD1B6.png" width="400" height="400">
However, I'm too lazy to look through all 10 rows and all 10 columns to figure out all the clues.
Could you please write me a program that can convert a drawn nonogram (represented as tuple of tuples):
```
nonogram = (
(0, 0, 0, 1, 0, 0, 0, 1, 1, 0),
(0, 0, 1, 1, 1, 0, 1, 1, 1, 1),
(0, 0, 1, 1, 1, 1, 1, 1, 1, 1),
(0, 0, 0, 1, 1, 1, 1, 1, 1, 0),
(0, 0, 0, 0, 0, 1, 1, 0, 0, 0),
(0, 1, 0, 0, 0, 0, 1, 1, 0, 0),
(1, 0, 1, 0, 0, 0, 1, 1, 0, 0),
(1, 1, 1, 0, 0, 1, 1, 0, 0, 0),
(1, 1, 1, 0, 0, 1, 1, 1, 0, 1),
(1, 1, 1, 1, 1, 1, 1, 1, 1, 1)
)
```
Into a tuple of such structure:
`(column_clues, row_clues)`
where
`column_clues = ((4,), (1, 3), (2, 4),...)` and `row_clues = ((1, 2), (3, 4), (8,),...)`
Notes:
1. Any empty rows or empty columns (which are cosmically unlikely to occur in random tests) should be represented as an emply (zero-length) tuple
2. Nonograms come in different sizes, such that `5 <= size <= 100`
3. There are 200 random tests
| algorithms | import itertools
def pack(m): return tuple(tuple(sum(g)
for k, g in itertools . groupby(r) if k) for r in m)
def encode(nonogram): return (pack(zip(* nonogram)), pack(nonogram))
| Multisize Nonogram Encoder | 629049687438580064f0e6dd | [
"Games",
"Algorithms"
] | https://www.codewars.com/kata/629049687438580064f0e6dd | 5 kyu |
A runner, who runs with base speed ```s``` with duration ```t``` will cover a distances ```d```: ```d = s * t```<br>
However, this runner can sprint for one unit of time with double speed ```s * 2```<br>
After sprinting, base speed ```s``` will permanently reduced by ```1```, and for next one unit of time runner will enter recovery phase and can't sprint again.
Your task, given base speed ```s``` and time ```t```, is to find the maximum possible distance ```d```.
## Input:
```1 <= s < 1000```<br>
```1 <= t < 1000```
## Example:
Given ```s = 2``` and ```t = 4```.<br>
We could schedule when runner should sprint so we could get these possible sequences:<br>
<pre>
Seq.: RRRR
=> s + s + s + s
=> 2 + 2 + 2 + 2 = 8
Seq.: RRRS
=> s + s + s + s*2
=> 2 + 2 + 2 + 2*2 = 10
Seq.: RRSR
=> s + s + s*2 + (s-1)
=> 2 + 2 + 2*2 + (2-1) = 9
Seq.: RSRR
=> s + s*2 + (s-1) + (s-1)
=> 2 + 2*2 + (2-1) + (2-1) = 8
Seq.: RSRS
=> s + s*2 + (s-1) + (s-1)*2
=> 2 + 2*2 + (2-1) + (2-1)*2 = 9
Seq.: SRRR
=> s*2 + (s-1) + (s-1) + (s-1)
=> 2*2 + (2-1) + (2-1) + (2-1) = 7
Seq.: SRRS
=> s*2 + (s-1) + (s-1) + (s-1)*2
=> 2*2 + (2-1) + (2-1) + (2-1)*2 = 8
Seq.: SRSR
=> s*2 + (s-1) + (s-1)*2 + (s-1-1)
=> 2*2 + (2-1) + (2-1)*2 + (2-1-1) = 7
Where:
- R: Normal Run / Recovery
- S: Sprint
</pre>
Based on above sequences, the maximum possible distance ```d``` is ```10```.
<p style="font-size: larger; text-align: center;"><b>Enjoy!</b></p> | algorithms | def solution(s, t):
# Sprints are optimally placed at the end of the race
'''
It is always beneficial to sprint at the end of the race
Sprints are optimally placed at the end of the race
e.g. RRRR...(SR)[n]S
We take a run without sprints are keep adding sprints at
the latest possible point until adding a sprint stops
increasing the total distance travelled
We can treat the final sprint as an add-on and count the number of
(SR) units we introduce into the race. Looking at this, each add
on unit changes the distance from 2*s to (3*s-3*n), so altering
the distance by (s-3*n). We find the largest n<s for which (s-3*n)>0
to find the trade-off point. If this is longer than the race in general
we do as many (SR) as possible.
'''
trade_off = s / / 3
# trade off must be less than max available slots
best_n_rs = min((t - 1) / / 2, trade_off)
def d(n): return s * t + s * (n + 1) - 3 / 2 * n * (n + 1)
return d(best_n_rs)
| Chaser's schedule | 628df6b29070907ecb3c2d83 | [
"Logic",
"Algorithms"
] | https://www.codewars.com/kata/628df6b29070907ecb3c2d83 | 6 kyu |
## About ##
In this task, you will write a "determinateValue" function that will determine the value of an element in a matrix.
The matrix should be filled with a zigzag, example matrix 5x5:
```
[
[ 1, 2, 6, 7, 15],
[ 3, 5, 8, 14, 16],
[ 4, 9, 13, 17, 22],
[10, 12, 18, 21, 23],
[11, 19, 20, 24, 25]
]
```
In example above if x = 2, y = 1 and n = 5, result should be 9 ( matrix[2][1] = 9 )
The function "determinateValue" will be called to determine the value of each element in the matrix, it is passed three parameters:
- x - index of array position (vertical index).
- y - index of element position in array (horizontal index).
- n - Matrix size.
Simple "code golf" conditions:
- C# - 250 chars limit.
- JS - 150 chars limit.
- Python - 160 chars limit.
## Note ##
- Matrices can be very large.
- Values must not go beyond the matrix (filling in the form of a big triangle will not work).
The direction of filling, for clarity.
<img 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" />
[co-author](https://www.codewars.com/users/Daniil%20Egorov) | algorithms | def determinateValue(x, y, n): return 1 + [y, x][(q := x + y) % 2] + q * (q + 1) / / 2 - max(q - n + 1, 0) * * 2
| [Code Golf] Fill matrix with a zigzag | 628d253eb110f3270a8a1789 | [
"Matrix",
"Algorithms",
"Restricted"
] | https://www.codewars.com/kata/628d253eb110f3270a8a1789 | 5 kyu |
## Overview
In this kata, `n` grains of rice will be initially placed at `n` **distinct** squares on a 2-dimensional rectangular board.
For example, if a particular grain starts on the board square at `x=3` and `y=8` it will be given starting "integer coordinates" `(3,8)`.
To keep things simple, all grains will always have coordinates `x>0` and `y>0`.
From the starting configuration of `n` grains in `n` distinct locations, your goal is to move the `n` grains individually so that they **all end up together on a same, shared, "finishing square".**
You must move each grain individually and you can only move each grain **1 square at a time** horizontally or vertically, but not diagonally.
Now, depending on the finishing square you choose, consider the **total number of moves**, M<sub>total</sub>, you take to move all `n` grains to that particular finishing square:
```
M_total(chosen finishing square) =
moves taken by grain 1 from its initial position to the finishing square +
moves taken by grain 2 from its initial position to the finishing square +
... +
moves taken by grain n from its initial position to the finishing square
```
Depending on which square of the board you choose to move all the `n` grains to, you will require a different **total number of moves**, M<sub>total</sub> .
**Which square(s) on the board requires the minimum total number of moves to move all `n` grains to it?**
---
## Inputs
- You will be given an `n`-element set of `2-tuples`, representing the starting integer coordinates of each of the `n` grains on the board.
- Each `2-tuple` will be of the form `(x,y)` with `x` and `y` both integers and both `> 0`.
- There will be no repeated `2-tuples`, corresponding to the fact that all `n` grains start at `n` distinct locations.
- The input provided to you will always be non-empty, i.e. will always contain at least 1 tuple. Your solution should be able to handle the case of `n=1` grains however.
**Performance requirements:** Values up to `n = 200,000` grains on a single board will be tested.
## Output
- You must return an array of `2-tuples` representing **all the possible finishing squares that can be reached in the minimum total number of moves** for the `n` grains. You can return the array in any order you like.
(Note that in some cases there may be **more than one square** in the board that corresponds to the minimum total number of moves, and you must find **all of them** for your solution to be valid).
---
## Walk-through Example
<a href="https://ibb.co/B61Pwn6"><img src="https://i.ibb.co/wQvLYcQ/ricegrains.png" alt="ricegrains" border="0"></a>
Here we have `n = 3` grains with their starting positions' integer coordinates given by:
`grains = {(1,3), (2,1), (4,3)}`
The **minimum total number of moves** to get all `n=3` grains together on the same square is M<sub>total</sub> = 5.
For this optimal value of M<sub>total</sub> = 5, there is only **1 possible finishing square**:
Finish square = `(2,3)` which corresponds to:
- grain #1 moving `1` step to the right
- grain #2 moving `2` steps up
- grain #3 moving `2` steps to the left.
If you try to move the `n=3` grains to any other possible finish square - such as, say, `(3,1)` - you will require more than M<sub>total</sub> = 5 total moves to do so.
For example, for `(3,1)`, it would take grain #1 `4` steps to reach it, grain #2 `1` step to reach it, and grain #3 `3` steps, for a total of `4+1+3 = 8` steps which is indeed greater than M<sub>total</sub> = 5 that we claimed is the minimum possible value.
For this example of input `grains = {(1,3), (2,1), (4,3)}` your solution should therefore return the array with one 2-element tuple: `[ (2,3) ]` | reference | # Return a list of 2-tuples of integers, representing
# the integer coordinates of ALL possible squares that
# satisfy the minimum number of total moves requirement
from statistics import median_low, median_high
def move_grains(grains):
xcoords = [x for x, _ in grains]
ycoords = [y for _, y in grains]
return [(x, y) for x in range(median_low(xcoords), median_high(xcoords) + 1) for y in range(median_low(ycoords), median_high(ycoords) + 1)]
| Move all the grains of rice to a same shared final square | 628d1c71b091df00651f0d83 | [
"Algorithms",
"Puzzles",
"Mathematics",
"Performance"
] | https://www.codewars.com/kata/628d1c71b091df00651f0d83 | 6 kyu |
## Overview
Start with a row of empty boxes, and place `n` bubbles in the left-most box in the row. You may assume there are infinitely many boxes to the right.
At each step, you must pick any box with 2 or more bubbles currently inside it, **merge 2 of the available bubbles in that box** to form a new bubble (thereby reducing the total number
of bubbles in the entire row by 1), then you must immediately **move the new bubble rightwards** into the **immediate next neighbour** box to the right.
You must continue doing this until you reach the final state in which you cannot merge any more bubbles; i.e. reach the state in which all boxes have **at most one** bubble in them.
For a given value of `n` initial bubbles placed in the left-most box in the row, **what is the number of steps required to reach the final state?**
(Hint: In case you are wondering why we can ask for **the** number of steps, and **the** final state, regardless of how you choose your steps: the number of steps for a given value of `n` will always be the same no matter how you choose your steps - see the Walk-through Example below)
---
## Inputs
You will be given an `integer`, `n`, representing the number of bubbles, which are all placed initially in the left-most box.
**Performance requirement:** You will be tested on values of `n` up to `200,000,000`.
## Outputs
Return an `integer`, representing the **number of steps required to reach the final state**.
---
## Walk-through Examples
```python
n = 10 bubbles
10 _ _ _ _
8 1 _ _ _
6 2 _ _ _
```
Note here that you can choose to either form `4 3 _ _ _` or `6 _ 1 _ _ _` let's pick the latter for this example:
```python
6 _ 1 _ _
4 1 1 _ _
2 2 1 _ _
```
Note here again that you can choose to either form `_ 3 1 _ _` or `2 _ 2 _ _` let's pick the former for this example:
```python
_ 3 1 _ _
_ 1 2 _ _
_ 1 _ 1 _ #final state
```
In this example we reached the valid final state - where all boxes contain at most one bubble - in `8` steps.
You can see in the above example that there may be more than one allowed move at any given step; for example, what if
we had picked option `4 3 _ _ _` on the 2nd step?
```python
n = 10 bubbles, alternative choices
10 _ _ _ _
8 1 _ _ _
6 2 _ _ _
4 3 _ _ _
```
Here again there are multiple possible legal choices, let's pick one:
```python
2 4 _ _ _
```
Again, multiple possible legal choices, let's pick one:
```python
_ 5 _ _ _
_ 3 1 _ _
_ 1 2 _ _
_ 1 _ 1 _ #final state
```
Note how, with these alternative choices, we have reached the same final state as in previous example **and in the exact same number of steps = 8** as in previous example. | reference | # The final state is the binary representation of
# the input, n, in reverse i.e. reading from left
# to right gives lsb to msb.
# So number of bubbles in final state in box i
# is equal to b_i from binary repr of n.
# To obtain 1 bubble in box i requires:
# S(i) = 2*S(i-1) + 1 steps with S(0)=0
# Solving the recurrence gives:
# S(i) = 2**i - 1
# Hence number of steps in total to place all bubbles
# is sigma{ b_i * S(i) } over all binary digits of n
# = sigma{ b_i * 2**i } - sigma{ b_i }
# = n - sigma{bits in n}
def moving_bubbles(n):
return n - n . bit_count()
| Merging and moving bubbles along a row of boxes | 628bd39474087000456cd126 | [
"Fundamentals",
"Algorithms",
"Puzzles",
"Mathematics"
] | https://www.codewars.com/kata/628bd39474087000456cd126 | 6 kyu |
Anna and Bob are playing tennis, they want you to show the result of their game.
A game is won when a player scores four (or more) points: 15, 30, 40 and the game-winning point.
Should both players make it 40, then the score is called "DEUCE". Following deuce, a player must win two consecutive points: the first point, known as "advantage" and the game-winning point.
*IMPORTANT : If the opposing player scores the next point when the player has the advantage, the game once again heads to deuce.
- 1 point = 15
- 2 points = 30
- 3 points = 40
- 40-40 = Deuce
- Server wins deuce point = Ad-In
- Receiver wins deuce point = Ad-Out
Array wins is given like following : wins = ["Anna","Bob","Anna"] (should return "Bob 15, Anna 30", Bob always goes first). Note that wins is never an empty array.
Write a function get_status(wins) which returns the result as a string as following :
- "DEUCE" when there is Deuce (40-40, or like said above)
- "30a" when both players have 30 and "15a" when both players have 15
- "Anna ADVANTAGE" when there is Deuce and Anna (for example) has scored
- "Bob WINS" when Bob (for example) has the advantage and scored
- or simply "Bob 15, Anna 30" for other cases
*NOTE : When a player has won, even if the other player has scored after, the winner is the winner (ex: get_status(["Bob","Anna","Bob","Anna","Bob","Anna","Bob","Bob","Anna"]) should return "Bob WINS")
| algorithms | def get_status(wins):
A = B = 0
for n in wins:
if n == "Anna":
A += 1
else:
B += 1
if (A > 3 or B > 3) and abs(A - B) > 1:
return f" { n } WINS"
return ("Anna ADVANTAGE" if B + 1 == A > 3 else
"Bob ADVANTAGE" if A + 1 == B > 3 else
["0a", "15a", "30a", "DEUCE"][min(A, 3)] if A == B else
f"Bob {[ 0 , 15 , 30 , 40 ][ B ]} , Anna {[ 0 , 15 , 30 , 40 ][ A ]} ")
| Give the status of the tennis game | 628b60fcb7d0770ddea8877d | [
"Games",
"Algorithms"
] | https://www.codewars.com/kata/628b60fcb7d0770ddea8877d | 6 kyu |
## OBJECTIVE:
You need to cross a certain section of the field in order to reach your target, but some cells of this field are irradiated with radiation, and you would not want to receive a large dose of radiation after completing your route.
The purpose of this kata is to create a function that takes a list of 10x10 cells (string type). On some of the cells there are radiation sources, which are of three types: <b><font color="orange">"A"</font></b>, <b><font color="orange">"B"</font></b>, <b><font color="orange">"C"</font></b>. It is necessary to calculate the degree of contamination of each free cell (by default, all cells except radiation sources have a radiation level of "0") and return a list with the correct values for each cell.
## DETAILS:
Each type of radiation source has its own characteristics of <b>radiation propagation</b> and <b>radiation power</b>:<br>
<b><font color="orange">Type "A"</font></b> has the following feature (<b>power=1</b>):
```
0 0 0 0 0
0 1 1 1 0
0 1 A 1 0
0 1 1 1 0
0 0 0 0 0
```
<b><font color="orange">Type "B"</font></b> (<b>power=2</b>):
```
0 0 0 0 0 0 0
0 0 0 2 0 0 0
0 0 0 2 0 0 0
0 2 2 B 2 2 0
0 0 0 2 0 0 0
0 0 0 2 0 0 0
0 0 0 0 0 0 0
```
<b><font color="orange">Type "C"</font></b> (<b>power=3</b>):
```
0 0 0 0 0 0 0
0 3 0 0 0 3 0
0 0 3 0 3 0 0
0 0 0 C 0 0 0
0 0 3 0 3 0 0
0 3 0 0 0 3 0
0 0 0 0 0 0 0
```
You need to take into account that when several sources act on one cell, the degree of its exposure should add up. For example, if a cell is affected by source "A" and source "B", then its degree will be equal to: (1+2)=3, and source "B" and "C": (2+3)=5.<br>
<b><font color="orange">For example:</font></b><br>
```
0 0 0 0 0 0 0 0 0 0
0 0 0 2 0 0 0 0 0 0
0 0 0 2 0 0 0 0 0 0
0 2 5 B 2 2 3 0 0 0
0 0 0 5 0 3 0 0 0 0
0 1 1 3 C 0 0 0 0 0
0 1 A 4 0 3 0 0 0 0
0 1 4 1 0 0 3 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
```
Sources <b>do not affect</b> each other in any way:
```
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 2 0 0 0 0
0 0 0 4 1 3 0 3 0 0
0 0 0 3 A B 5 2 0 0
0 0 0 1 1 C 0 0 0 0
0 0 0 0 3 2 3 0 0 0
0 0 0 3 0 0 0 3 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
```
If the radiation source is in such a position that some of its rays will not be displayed on the field, mark only those that are possible.<br>
<b><font color="orange">For example:</font></b><br>
```
A 1 0 0 0
1 1 0 0 0
0 0 0 0 0
0 0 0 0 0
``` | games | from copy import deepcopy
RADIATION_SOURCES = {
# You can easily modify:
# symbol: ((x, y, val), ...), # affected cells and impact value
'A': (
(- 1, - 1, 1), (0, - 1, 1), (1, - 1, 1),
(- 1, 0, 1), (1, 0, 1),
(- 1, 1, 1), (0, 1, 1), (1, 1, 1),
),
'B': (
(0, - 2, 2),
(0, - 1, 2),
(- 2, 0, 2), (- 1, 0, 2), (1, 0, 2), (2, 0, 2),
(0, 1, 2),
(0, 2, 2),
),
'C': (
(- 2, - 2, 3), (2, - 2, 3),
(- 1, - 1, 3), (1, - 1, 3),
(- 1, 1, 3), (1, 1, 3),
(- 2, 2, 3), (2, 2, 3),
),
}
def calculate_cells(field):
cp_field = deepcopy(field)
width = (height := len(cp_field)) and len(cp_field[0])
for y in range(height):
for x in range(width):
for a, b, val in RADIATION_SOURCES . get(cell := cp_field[y][x], []):
if 0 <= (ny := y + b) < height and 0 <= (nx := x + a) < width:
if (nv := cp_field[ny][nx]) not in RADIATION_SOURCES:
cp_field[ny][nx] = str(int(nv) + val)
return cp_field
| Exposure assessment | 628b633fb7d0770d57a88809 | [
"Algorithms",
"Fundamentals"
] | https://www.codewars.com/kata/628b633fb7d0770d57a88809 | 6 kyu |
### Introduction
An IPv4 address is made up of 32 bits. The bits are divided into four sections called **octets** (groups of eight). Each octet can have a number in the range 0 to 255 (unsigned 8-bit integers).
All IPv4 addresses belong to an **Address Class**. There are five different address classes (A, B, C, D and E). It's the **first** (left most) octet that tells us which class the address belongs to.
```
Class: Identifier: Ranges:
---------------------------------------
A 0------- 00000000-01111111
B 10------ 10000000-10111111
C 110----- 11000000-11011111
D 1110---- 11100000-11101111
E 1111---- 11110000-11111111
```
The bits in the **Identifiers** column above is used to determine the class of an address. So, if an address starts with the bits that are visible in the identifier column, it belongs to the class in the **Class** column on the same row.
If we look at the address `192.168.0.1` as an example:
* The first octet is `192`, which in binary is `11000000`.
* The first 3 bits of this octet is set to `110` which means that it is a **Class C** address.
### The Assignment
~~~if:c
Your task is to implement a function `char ipv4_address_class(const char *ipv4_addr)` which takes in an IPv4 address `ipv4_addr` and returns a character `char` (A, B, C, D, or E) representing the address class.
~~~
~~~if:python
Your task is to implement a function `def ipv4_address_class(ipv4_addr` which takes in an IPv4 address string `ipv4_addr` and returns a single character string (`'A'`, `'B'`, `'C'`, `'D'`, or `'E'`) representing the address class.
~~~
### Testing
The tests will challenge your function with addresses in the full 32-bit range (`0.0.0.0 - 255.255.255.255`) and ***nothing else***.
### Links
Check out these links for more information:
* https://en.wikipedia.org/wiki/Classful_network
* https://en.wikipedia.org/wiki/Binary_number
| reference | def ipv4_address_class(ipv4_addr):
n = int(ipv4_addr . split(".")[0])
return " EDCBA" [(n <= 127) + (n <= 191) + (n <= 223) + (n <= 239) + (n <= 255)]
| Networking 101/2: Identify the IPv4 address Class | 628ba76a85a2d500649da696 | [
"Networks",
"Bits",
"Fundamentals",
"Binary"
] | https://www.codewars.com/kata/628ba76a85a2d500649da696 | 7 kyu |
### Introduction
Both the IPv4 address and subnet mask is made up of 32 bits. The bits are divided into four sections called octets (groups of eight). Each octet can have a number in the range 0 to 255 (unsigned 8-bit integers).
When learning IPv4 addressing it is essential to know how the Binary Number System works as well as having the right tools for the task you are facing. But, you are lacking a good peace of code to visualize the IPv4 addresses and subnet masks in binary form...
### The Assignment
~~~if:c
Your task is to implement a function `void ipv4_to_binary(const char *ipv4_addr, char ipv4_binary[36])` which takes in an IPv4 address `ipv4_addr` and writes the result converted into **Dotted Binary Notation** in the output parameter `ipv4_binary`, a pre-allocated character buffer large enough to hold the result. The function has no return value.
~~~
~~~if:python
Your task is to implement a funtion `def ipv4_to_binary(ipv4_addr: str) -> str`
which takes in an IPv4 address `ipv4_addr`, converts it into **Dotted Binary Notation** and returns it as a string.
~~~
Here are a few examples of input (left) and output (right):
192.168.0.33 = 11000000.10101000.00000000.00100001
255.255.255.0 = 11111111.11111111.11111111.00000000
10.55.0.13 = 00001010.00110111.00000000.00001101
255.255.255.252 = 11111111.11111111.11111111.11111100
`192.168.0.33` in the example above is the **Dotted Decimal Notation** and
`11000000.10101000.00000000.00100001` is the corresponding **Dotted Binary Notation**.
>Having a function that prints an address this way might make it easier when making various *bitwise* calculations later on (in subsequent katas).
### Testing
The tests will challenge your function with addresses in the full 32-bit range (`0.0.0.0 - 255.255.255.255`) and ***nothing else***.
### Links
Check out these links for more information:
* https://en.wikipedia.org/wiki/IPv4#Addressing
* https://en.wikipedia.org/wiki/Binary_number
| reference | def ipv4_to_binary(ipv4_addr: str) - > str:
return "." . join(f" { octet :0 8 b } " for octet in map(int, ipv4_addr . split(".")))
| Networking 101/1: Convert IPv4 address to Dotted Binary Notation | 6288de23ab7ede0031602521 | [
"Networks",
"Bits",
"Fundamentals",
"Binary"
] | https://www.codewars.com/kata/6288de23ab7ede0031602521 | 6 kyu |
## Overview
Given an alphabet consisting of `a` distinct symbols, how many strings of exact length = `n` are there which use **all** of the `a` distinct symbols **at least once?**
Equivalent reformulation, if you prefer: how many strings are not missing **any** of the `a` distinct symbols?
For example, given the alphabet `{F,G}` consisting of `a=2` distinct symbols, the string of length `n=5` given by `FGGFG` is a permitted string because it uses all of the `a` symbols at least once. On the other hand, the string `FFFFF` is a forbidden string because it does **not** use all of the symbols from the alphabet `{F,G}` at least once (it doesn't use the symbol `G`).
---
## Inputs
You will be given two `integers`, `n` and `a`, representing the length of the string, and the size of the alphabet i.e. the number of symbols.
**Performance requirement:** In the tests, `n` and `a` will be tested together up to `n=250` and `a=150`. Therefore, an approach in which you attempt to generate all possible strings and count which ones satisfy the requirement will probably time out. Instead, use the walk-through examples below to develop some insight.
## Outputs
Return an `integer`, representing the number of strings of length `n` that use all of the symbols from an alphabet of size `a` at least once.
---
## Walk-through examples
-**Example 1:** `n = 5` and `a = 2`
Suppose we use the alphabet of 2 symbols: `{F,G}` (the specific symbols `F` and `G` don't matter, of course, just that there are `a = 2` of them).
There are `2**5` possible strings of length `n = 5` using these 2 possible symbols.
Among all `2**5` such possible strings, one of them is the forbidden string `FFFFF` which **DOES NOT** contain the symbol `G` at least once, and one of them is the forbidden string `GGGGG` which **DOES NOT** contain the symbol `F` at least once.
In all of the other possible strings, **all** the symbols in the alphabet `{F,G}` appear at least once.
Hence there are `2**5 - 1 - 1` = `2**5 - 2` = `30` strings of exact length = `n = 5` in which all of the `a = 2` distinct symbols appear at least once.
-**Example 2:** `n = 5` and `a = 3`
Again, suppose we use the alphabet of 3 symbols: `{F,G,H}`.
There are `3**5` possible strings of length `n = 5` using these 3 possible symbols.
Let's count the forbidden strings according to the **exact** number of symbols that they do not use.
Firstly, there are some forbidden strings missing **exactly 2 of the 3 symbols**: among all `3**5` possible strings, one of them is the string `FFFFF` which **DOES NOT** contain either the symbols `G` or `H` at least once, one of them is the string `GGGGG` which **DOES NOT** contain either the symbols `F` or `H` at least once, and one of them is the string `HHHHH` which **DOES NOT** contain either the symbols `F` or `G` at least once. Thus there are in total `1 + 1 + 1 = 3 * 1 = 3` forbidden strings which are missing **exactly 2** of the `a = 3` symbols in the alphabet.
Now that we have considered the forbidden strings that are missing **exactly 2 of the 3 symbols**, we realise that there are also forbidden strings which are missing **exactly 1** of the `a = 3` symbols in the alphabet. For example, how many forbidden strings are there that are missing **exactly** the symbol `H` ? i.e. strings which **are not missing** the symbols `F` **or** `G` ?
The answer is quite simply the number of strings with `n = 5` and `a = 3-1 = 2` in which all of the `a = 3-1 = 2` remaining symbols **must appear at least once**. But this is **exactly what we calculated in Example 1.**
So, **for each of the 3 choices** of `F, G, or H` there are exactly `2**5 - 2 = 30` forbidden strings in which **exactly one** of those symbols is missing, **but not two of them**.
So, among all `3**5` strings, there are `3 * 1 = 3` forbidden strings missing exactly 2 of the symbols, and there are `3 * ( 2**5 - 2)` forbidden strings missing exactly 1 symbol.
Thus there are `3**5 - 3 - 3*(2**5 - 2) = 150` strings which are missing exactly 0 symbols, i.e. `150` strings in which all of the `a = 3` symbols appear at least once.
---
## Hint
If you'd like to check the above results, for these small values of `n` and `a` you can easily generate all `2**5` and all `3**5` strings and loop over them, counting which ones have all of the symbols appearing at least once.
This approach will not work for the larger values of `n` and `a` used in the full tests though, since you won't be able to generate all `150**250` strings when e.g. `n = 250` and `a = 150`, but hopefully the walk-through examples above have revealed a pattern that you can use. | reference | from math import comb
def use_all_symbols(n, a):
return sum((- 1) * * i * comb(a, i) * (a - i) * * n for i in range(a + 1))
| How many strings use all symbols of a given alphabet at least once | 626ec08b40a15e2d250575cf | [
"Algorithms",
"Mathematics"
] | https://www.codewars.com/kata/626ec08b40a15e2d250575cf | 5 kyu |
Given the name of a [US state or territory](https://en.wikipedia.org/wiki/List_of_states_and_territories_of_the_United_States), return its postal abbreviation. All states, federal districts and inhabited territories will be tested, according to the linked wikipedia page.
***Notes:***
* to spark your creativity, the size of your code is limited to 500 characters
* the list of states and their respective abbreviations are listed below
```if:python
* `import`, `exec`, `eval` and `"__"` are forbidden
```
```if:ruby
* `require`, `exec`, and `eval` are forbidden
```
### Examples
```python
"Alabama" --> "AL"
"District of Columbia" --> "DC"
"U.S. Virgin Islands" --> "VI"
```
### List of states and abbreviations
```
Alabama, AL
Alaska, AK
Arizona, AZ
Arkansas, AR
California, CA
Colorado, CO
Connecticut, CT
Delaware, DE
Florida, FL
Georgia, GA
Hawaii, HI
Idaho, ID
Illinois, IL
Indiana, IN
Iowa, IA
Kansas, KS
Kentucky, KY
Louisiana, LA
Maine, ME
Maryland, MD
Massachusetts, MA
Michigan, MI
Minnesota, MN
Mississippi, MS
Missouri, MO
Montana, MT
Nebraska, NE
Nevada, NV
New Hampshire, NH
New Jersey, NJ
New Mexico, NM
New York, NY
North Carolina, NC
North Dakota, ND
Ohio, OH
Oklahoma, OK
Oregon, OR
Pennsylvania, PA
Rhode Island, RI
South Carolina, SC
South Dakota, SD
Tennessee, TN
Texas, TX
Utah, UT
Vermont, VT
Virginia, VA
Washington, WA
West Virginia, WV
Wisconsin, WI
Wyoming, WY
District of Columbia, DC
American Samoa, AS
Guam, GU
Northern Mariana Islands, MP
Puerto Rico, PR
U.S. Virgin Islands, VI
```
---
### My other katas
If you enjoyed this kata then please try [my other katas](https://www.codewars.com/users/anter69/authored)! :-)
#### *Translations are welcome!*
| algorithms | N = "Alab,Alas,Am,Ari,Ark,Ca,Col,Con,De,Di,F,Ge,Gu,H,Id,Il,In,Io,Ka,Ke,L,Mai,Mar,Mas,Mic,Min,Missi,Misso,Mo,Neb,Nev,New H,New J,New M,New Y,North C,North D,Northe,Oh,Ok,Or,Pe,Pu,R,South C,South D,Ten,Tex,U.,Ut,Ve,Vi,Wa,We,Wi,Wy" . split(
",")
C = "AL AK AS AZ AR CA CO CT DE DC FL GA GU HI ID IL IN IA KS KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC ND MP OH OK OR PA PR RI SC SD TN TX VI UT VT VA WA WV WI WY" . split()
def abbr(s):
return next(c for n, c in zip(N, C) if s . startswith(n))
| US Postal Codes | 60245d013b9cda0008f4da8e | [
"Algorithms",
"Restricted"
] | https://www.codewars.com/kata/60245d013b9cda0008f4da8e | 5 kyu |
## Overview
`n` very naughty Codewars users want to cheat amongst themselves, and share solutions to katas they have solved.
Each of the `n` users was only clever enough to solve 1 distinct kata on their own, and now they would like to know the solutions to **each other's unique kata** that each of the other `n - 1` users was able to solve.
**Any two** naughty Codewars users can chat to eachother on Discord, and during their naughty cheater **2-person conversation** they will **exchange ALL the solutions they currently know with the other user**.
The `n` naughty Codewars users now want to work out what the optimal way to organize their Discord chats is.
In other words, they want to **find a sequence of 2-person conversations that reaches a final state where all `n` users know all `n` solutions, in the smallest possible number of conversations.**
Since they are all noobs, they have asked you to solve this problem for them!
*Disclaimer: rest assured, no cheaters will be helped by you solving this kata. Any resemblance to actual Codewars users is purely coincidental. There is no secret cheater Discord.*
---
## Example with `n = 3` Codewars users
```
these 2 users
will have a
conversation -> (1,3) -> (2,3) -> (1,2)
_____________ _______ _________ ________
user 1: {1} {1,3} {1,3} {1,2,3}
user 2: {2} {2} {1,2,3} {1,2,3}
user 3: {3} {1,3} {1,2,3} {1,2,3}
```
The above illustration shows how the series of conversations `[(1,3), (2,3), (1,2)]` gradually leads to all 3 users gradually learning the solutions to all the different katas.
Initially, all 3 users only know 1 solution each (user 1 knows solution to kata `{1}`, etc.)
Then, for example, when user 1 and user 3 have their first conversation, they exchange all their known solutions, so after the first conversation `(1,3)` we have:
`user 1 knows: {1,3}, user 2 knows: {2}, user 3 knows: {1,3}`
Note that **there are many possible ways of reaching the desired final state**, for example you can check that `[(1,2), (3,2), (1,2)]` also works, but they all require at least 3 conversations. It is **not** possible to do this with a smaller number of conversations, so the **optimal length** is therefore 3.
---
## Inputs
You will be given an `integer` `n` representing the number of naughty Codewars users.
The input `n` will always be `n >= 2`.
**Performance requirements:** Values of `n` up to `n = 1000` will be used in tests.
## Outputs
Return an array of 2-element `tuples` of `integers`, representing the 2-person conversations that should take place, and with the **array sorted in the order in which the conversations should take place**.
The length of this array will therefore be the **number of conversations** that your algorithm determines to be the optimal i.e. fewest possible.
The order of the 2 users in each of the 2-element tuples **does not** matter: the solution checker will treat `(user_1, user_2)` as the same conversation as `(user_2, user_1)`.
---
## Important note on tests
For a given `n`, the exact series of conversations leading to optimal solution is not unique, as illustrated in the Example for `n=3` above, but the **optimal number of conversations is well defined**.
Therefore, the tests will check **2** features of your solution:
1. Tests will first check that your returned array of 2-person conversations does indeed lead to a final state in which all `n` users have the solutions to all `n` katas.
1. Secondly, tests will then check for the OPTIMALITY of your proposed list of conversations; i.e. that its length is equal to the minimum possible for the given value of `n`.
**If you are failing tests despite returning a valid list of conversations** it is probably because your solution is passing the tests corresponding to requirement 1) but not passing the optimality requirement 2). In other words, your solution is leading to a valid final state, but it is not doing so in the **minimum possible** number of conversations. | reference | def optimal_conversations(n):
if n == 2:
return [(1, 2)]
if n == 3:
return [(1, 2), (1, 3), (2, 3)]
# Optimal solution for 4 people
res = [(1, 2), (3, 4), (1, 3), (2, 4)]
# If more than 4 people, just get their information at first and give them the (1, 2, 3, 4) at the end
above_4 = [(1, i) for i in range(5, n + 1)]
return above_4 + res + above_4
| Help naughty Codewars users cheat by sharing their solutions to katas | 627183ce5aa8580057592940 | [
"Puzzles",
"Mathematics",
"Logic",
"Combinatorics"
] | https://www.codewars.com/kata/627183ce5aa8580057592940 | 6 kyu |
Simpler and similar kata <a href=http://www.codewars.com/kata/multiperfect-number >here</a>.
A generalization of <a href=https://en.wikipedia.org/wiki/Perfect_number>perfect numbers</a> are (m,k)-perfect number.
It is a series of numbers that satisfy the following condition:
```math
\Huge
\sigma^{m}(n) = kn
```
where <font size=5>σ<sup>m</sup></font> is the divisor function (or sum of all divisors) <a href=https://en.wikipedia.org/wiki/Divisor_function>[Wikipedia link]</a> applied ```m``` times, ```k``` a constant.
For example "normal" perfect numbers are (1,2)-perfect number.
<font size=5>Goal:</font><br>
Write a function ```first_mk_perfectnumber(m,k)``` that, give in input the ```m``` and ```k``` of the previous equation ( σ<sup>m</sup>(n) = kn ) , it returns the first number (```n```) that satisfy the relation.
Pay attention to the <u>performance</u> of your algorithm!! This function must run within the maximum time very big number (>200000).
My solution's time of execution in any language is around 4 seconds.
Wikipedia Page : <a href=https://en.wikipedia.org/wiki/Superperfect_number>Superperfect Number [Generalization at bottom]</a>
<font size=5>Example:</font><br>
```first_mk_perfectnumber(2,2)``` return 2 because <br>
σ(σ(1)) = 1 so ```1 != 2*1```
while σ(σ(2)) = 4
so ```4 == 2*2```
<big><u>Hint 1</big>: You can execute divisor function speedly with prime factors.</u> You can find formula here: <a href=https://en.wikipedia.org/wiki/Divisor_function >Wiki Page</a> <a href=http://mathworld.wolfram.com/DivisorFunction.html#h-100 >Wolfram Page</a>
| algorithms | from functools import cache
from gmpy2 import is_prime, next_prime
@ cache
def σ(m, n):
s, p = 1, 2
while n > 1:
while n % p:
p = n if is_prime(n) else next_prime(p)
i = 1
while n % p == 0:
n / /= p
i += 1
s *= (p * * i - 1) / / (p - 1)
return s if m == 1 else σ(m - 1, s)
def first_mk_perfectnumber(m, k):
i = 1
while σ(m, i) != k * i:
i += 1
return i
| (m, k)-Perfect Number | 5597ca93d5b5b37db8000066 | [
"Mathematics",
"Performance",
"Algorithms"
] | https://www.codewars.com/kata/5597ca93d5b5b37db8000066 | 5 kyu |
You are a big fan of Formula 1.
There was a race last week-end, but ... you missed the live TV.
You picked on the internet a list of the **events** during the race, and you want to get your champion's rank at the end of the race.
An event is a pilot ID followed by a type. There are two types of events:
<ul><li>'O' : a pilot is overtaking the one just in front of him</li>
<li>'I' : a pilot had an incident... and is forced to quit the race (and subsequently all pilots after this one gain one rank)</li></ul>
The list of events is passed as a string where events are separated with a space, first event first.
_e.g: "2 O 12 I" means the pilot#2 overtook, then the pilot#12 had an incident._
Your function will return the rank of your champion at the end of the race, given your champion's ID (as int) and the list of events. Return -1 if your champion had an incident.
NOTES:
<ul><li>there are 20 positions when the race starts (from 1 to 20, 1 being the one in front),</li>
<li>therefore, the <strong>rank</strong> of a pilot, at start and during the race, is actually his/her <strong>position</strong> (first pilot at position 1, etc...),</li>
<li>each pilot ID is his position when the races starts,</li>
<li>all events in entry lists are syntactically valid.</li>
</ul>
_Final tests contains about 20 fixed events, and 100 random tests with race length between 10 to 30 events.
In the tests (fixed and random): the leading pilot never overtakes, and after having an incident, the pilot never appears in further events._
**Enjoy !**
_Thanks to the **Best french developer Contest** 2019_
| algorithms | import re
def champion_rank(pilot: int, events: str) - > int:
arr = [i for i in range(1, 21)]
for x in re . findall("\d+ [IO]", events):
r, t = x . split()
if t == 'I':
arr . remove(int(r))
else:
j = arr . index(int(r))
arr[j], arr[j - 1] = arr[j - 1], arr[j]
return arr . index(pilot) + 1 if pilot in arr else - 1
| Formula 1 Race | 626d691649cb3c7acd63457b | [
"Algorithms"
] | https://www.codewars.com/kata/626d691649cb3c7acd63457b | 5 kyu |
## Overview
`N` White frogs and `N` Black frogs are placed on either side of a line made up of `2*N + 1` spaces. The 1 "leftover" space is therefore empty, and is at the center of the line.
The `N` White frogs are placed on the left side and all face RIGHTWARDS, and the `N` Black frogs are placed on the right side all face LEFTWARDS.
Your goal is to get all the White frogs to the right half of the line, and all the Black frogs to the left half of the line.
`N=2, start: WW_BB --> BB_WW :end`
**A white frog must move first,** then you may choose from any of the 3 following moves, each encoded by a single letter `w`,`b`, or `j`:
- `w`: move a White frog one space rightwards, if that space is empty. `WWW_ -> WW_W`
- `b`: move a Black frog one space leftwards, if that space is empty. `_BBB -> B_BB`
- `j`: make a rightwards jump with a White frog, or a leftwards jump with a Black frog, over **ONE AND ONLY ONE frog of the opposite color**, and **only if the frog can land in the empty space**. `WB_B -> _BWB`
Can you develop a general algorithm for arbitrary `N` that moves all the frogs to their end positions in the fewest number of moves?
---
## Inputs and Outputs
You will be given an integer `N` which specifies the number of both types of frogs i.e. `N` White and `N` Black frogs. The board will therefore always be of size `2*N + 1`.
You must return a `string` of the letters `w`, `b`, `j`, which encodes the moves your algorithm takes to solve the problem for a given value of `N`.
**REMEMBER: a white frog must move first.** With this condition, there will always be a **unique optimal solution**, which will consist of `N**2 + 2*N` moves, for every value of `N`.
---
## Hint
You may be wondering how you can encode the different possible jump moves using only a single letter `j` - for example, if you somehow reached the configuration `WB_WB` there are 2 possible jump options: `WB_WB -> _BWWB` or `WB_WB -> WBBW_`, so how should you specify "which frog does the jump"?
Answer: in the optimal solution, every jump move will be **uniquely defined**.
So if your solution ever reaches a configuration where you find there are 2 or more possible jump options, you are guaranteed to eventually reach a dead-end, which means that your current list of moves will not lead to the correct answer!
---
## Examples
**N = 1, optimal solution**
```python
W_B # initial state
_WB # state after performing move: 'w'
BW_ # state after performing move: 'j'
B_W # state after performing move: 'w'
```
Solution string for `N = 1` is therefore: `'wjw'`. This solution is optimal and does indeed consist of `N**2 + 2*N = 1+2 = 3` moves.
**N = 2, optimal solution**
```python
WW_BB # initial state
W_WBB # state after performing move: 'w'
WBW_B # state after performing move: 'j'
WBWB_ # state after performing move: 'b'
WB_BW # state after performing move: 'j'
_BWBW # state after performing move: 'j'
B_WBW # state after performing move: 'b'
BBW_W # state after performing move: 'j'
BB_WW # state after performing move: 'w'
```
Solution string for `N = 2` is therefore: `'wjbjjbjw'`. This solution is optimal and does indeed consist of `N**2 + 2*N = 4+4 = 8` moves. | reference | def white_black_frogs(n):
return '' . join('j' * i + 'wb' [i & 1] for i in range(n)) + '' . join('j' * (n - i) + 'bw' [(i + n) & 1] for i in range(n))
| Help the frogs reach the opposite side of the line | 626d96eb49cb3c7a2f634bbf | [
"Algorithms",
"Puzzles",
"Recursion"
] | https://www.codewars.com/kata/626d96eb49cb3c7a2f634bbf | 6 kyu |
## Overview
In this kata you will be working on a 3x3 chessboard. In each test case you will be given:
- A **starting 3x3 board configuration** with some White and Black Knights on the 3x3 board.
- An **ending 3x3 board configuration** with some White and Black Knights.
Your goal is to determine if you can **reach the end board configuration from the start board configuration**.
This is a simple puzzle version of Knight movement, so some rules are different compared to real chess:
**Firstly, you may not capture pieces** - so for example a White knight **can not** move to a square that's occupied by a Black knight and "remove" the piece. Therefore no pieces will be removed during each game.
**Secondly**, you may only move one Knight at a time, but - **unlike in real chess** - you may choose to move **any number** of White pieces without "swapping color" to Black, and vice versa for Black pieces. In other words, you can move 1,2,3,4,... White knights around the board before you move a single Black piece etc.
Finally, just like in standard chess moves, a Knight can move
2 horizontal and 1 vertical position, or 1 horizontal and 2 vertical positions.
---
## Inputs
You will be given 2 `lists` (Python), 1 `list` called `start_board` representing the 3x3 state in the starting configuration, and 1 `list` called `end_board` representing the ending configuration.
Each such `list` contains 3 `lists`, each of which consists of 3 `strings`, to represent an entire 1x3 row of the chessboard.
3x3 board layout:
```python
[ [top_left_square, top_middle_square, top_right_square],
[middle_left_square, middle_middle_square, middle_right_square],
[bottom_left_square, bottom_middle_square, bottom_right_square] ]
```
If a White Knight is on a given square, the corresponding location in the board will have an `'W'` string, if a Black Knight is on a square the location will have a `'B'` string. If a square is empty there will be a `'_'` string.
Example input list:
```python
start_board = [ ['B' , '_', '_' ], [ '_', '_', 'W'] , [ 'W', '_', '_'] ]
```
The above example represents a starting configuration where there is a Black Knight in the top_left_square, a White Knight in the middle_right_square, and a White Knight in the bottom_left_square.
## Outputs
Return `True` if it is possible to reach the given `end_board` from the given `start_board` using the legal moves.
Return `False` if it is **not** possible.
---
## Examples
Example 1
```python
start_board = [ ['W' , '_', '_' ], [ '_', '_', '_'] , [ '_', '_', '_'] ]
end_board = [ ['W' , '_', '_' ], [ '_', '_', '_'] , [ '_', '_', '_'] ]
```
Expected solution is `True`; there is only one White Knight and it doesn't need to move between start and end configuration.
Example 2
```python
start_board = [ ['W' , '_', '_' ], [ '_', '_', '_'] , [ '_', '_', '_'] ]
end_board = [ ['_' , '_', '_' ], [ '_', '_', '_'] , [ '_', 'W', '_'] ]
```
Expected solution is `True`; there is only one White Knight and it can reach the end configuration from the start configuration (in fact, it only requires one move to get there).
Example 3
```python
start_board = [ ['W' , '_', '_' ], [ '_', '_', '_'] , [ '_', '_', '_'] ]
end_board = [ ['B' , '_', '_' ], [ '_', '_', '_'] , [ '_', '_', '_'] ]
```
Expected solution is `False`; the start configuration has only one White Knight and the end configuration has only one Black Knight.
Example 4
```python
start_board = [ ['_' , 'B', '_' ], [ '_', '_', '_'] , [ '_', '_', 'W'] ]
end_board = [ ['_' , '_', '_' ], [ 'W', '_', 'B'] , [ '_', '_', '_'] ]
```
Expected solution is `True`; there is one White Knight and one Black Knight. It takes a minimum total of 1W+2B = 3 moves to get all the pieces from the start configuration to the end configuration. | reference | def is_reachable(s, e):
ss, es = ['' . join(b[y][x] for y, x in [(0, 0), (1, 2), (2, 0), (0, 1),
(2, 2), (1, 0), (0, 2), (2, 1)] if b[y][x] != '_') for b in (s, e)]
return s[1][1] == e[1][1] and ss in es * (1 + (len(es) < 8))
| Configurations of White and Black Knights on a 3x3 chessboard | 626b949b40a15ed6e7055b8f | [
"Fundamentals",
"Algorithms",
"Puzzles",
"Mathematics",
"Performance"
] | https://www.codewars.com/kata/626b949b40a15ed6e7055b8f | 5 kyu |
Wordle is a game where users try to guess a hidden 5 letter word. Feedback is given for each guess, in the form of colored tiles, indicating when letters match or occupy the correct position.
see: https://www.nytimes.com/games/wordle/index.html
### The Brief
Make a function that takes in a 5 letter guess and, comparing to a known solution, returns clues for the user to solve the wordle puzzle.
### Requirements
- The function takes in two strings of 5 letters each, a guess and a solution.
- The function returns a string of 5 letters.
- The returned string is composed of letters each corresponding to a letter of the guess: 'b' (black) when the letter is not present in the answer; 'y' (yellow) when the letter is present but in a different location; and 'g' (green) when the letter is present and in the right location.
### Duplicates
- A letter appearing in the correct place is always green, including if it appears elsewhere in the word.
- A letter can only be green or yellow as many times as it appears in the correct answer. Extra occurences of that letter are marked as 'b'. Below are some examples, see example tests for more:
- letters are evaluated left to right.
```
Guess: xxxxx
Answer: yxxxx
Expected Output: bgggg
Explanation: extra letters marked as 'b'
guess: xxxxz
answer:aaaax
expected output: ybbbb
first x marked as 'y', and each one as 'b' after
```
| algorithms | def resolver(guess, solution):
letters, answer = {}, ['b'] * len(guess)
for i, (g, s) in enumerate(zip(guess, solution)):
if g == s:
answer[i] = 'g'
else:
letters[s] = letters . get(s, 0) + 1
for i, (g, s) in enumerate(zip(guess, solution)):
if g != s and letters . get(g, 0) > 0:
answer[i] = 'y'
letters[g] -= 1
return '' . join(answer)
| A Brave New Wordle | 62013b174c72240016600e60 | [
"Algorithms",
"Strings",
"Puzzles"
] | https://www.codewars.com/kata/62013b174c72240016600e60 | 5 kyu |
## Overview
On an n x n chessboard, a standard King can move 1 square at a time in any of the following 8 possible directions:
- up / down / left / right = the 4 possible linear directions
- upleft / upright / downleft / downright = the 4 possible diagonal directions.
Imagine you now have a non-standard **Limited King** who can move 1 square at a time **in ONLY the following 3 possible directions**:
- down = linear
- right = linear
- upleft = diagonal
Can you create a **"Limited King's Tour"** of a standard n x n chessboard?
In other words: can you find a way to visit **EVERY SQUARE** on the n x n chessboard **EXACTLY ONCE**, using a Limited King who can only move in the 3 possible directions listed above?
---
## Important notes
1. You are allowed to **choose your starting square** for your Limited King's Tour on **any square you want** on the board.
2. Your Tour of the board does **NOT** have to be **closed** - i.e. once your Limited King reaches the final square on its Tour of the board, it **does not** have to be able to reach the initial square again (such Tours are called "closed" or "re-entrant" tours in the context of other chess pieces, if you want to learn more about them)
---
## Input
You will be given an integer, `n`, that specifies the size of the n x n chessboard to solve.
**Performance requirements:** The maximum value of `n` used in the random tests will be `n = 750` .
## Output
1. You must return a `list` (Python), representing your complete solution to the Limited King's Tour on the n x n board.
2. Each element in your solution `list` should be a 2-component`tuple` (Python) and will represent a square's location `(row,column)` on the board. The first `tuple` in your solution is therefore the square where your Limited King's Tour begins, and the last `tuple` is where your Tour ends.
3. **IMPORTANT INDEXING CONVENTION** You must use index numbers to represent squares, so for example `(0,0)` is the top-left square of the board, and `( n-1, n-1 )` is the bottom-right square of the n x n board.
4. **Hint: There is always at least one solution for any value of `n`. If your algorithm can find more than one valid solution, you should only return one of them.**
---
## Example
For `n = 2`, one possible valid solution to the Limited King's Tour is given by the following list: `[(1, 0), (1, 1), (0, 0), (0, 1)]` which corresponds to starting on bottom-left square of the board `(1,0)` then moving right to the bottom-right square of the board `(1,1)` etc. Note how all `n * n = 4` possible squares on the `n = 2` board have been used **exactly once**.
For `n = 3`, one possible valid solution is: `[(0,1),(0,2),(1,2),(2,2),(1,1),(0,0),(1,0),(2,0),(2,1)]`. Again, note how all `n * n = 9` possible squares on the `n = 3` board have been used exactly once.
---
## Tests
To help you understand error messages in case of a failed test:
The tests implement a solution checker that will return True if your solution is valid, and False if your solution is invalid for any of the following reasons:
-Your solution must be of length n x n (since there are n x n squares on the board for the Limited King to visit)
-Each possible square on the n x n board must appear **EXACTLY ONCE** in your solution
-Each `tuple` in position *"i+1"* in your solution list must be reachable from the `tuple` in position *"i"*, using one of the 3 allowed Limited King moves | reference | def limited_king(n):
res = n % 3 == 1 and [(0, j) for j in range(n)] or []
for x in range(bool(res), n, 3):
res . extend((k, l) for l in range(n - 1, - 1, - 1)
for k in range(x, x + 2))
if x + 2 != n:
res . extend((x + 2, j) for j in range(n))
return res
| Find a Limited King's Tour on a chessboard | 626868a414c0146ddbe0e6be | [
"Fundamentals",
"Algorithms",
"Puzzles",
"Mathematics"
] | https://www.codewars.com/kata/626868a414c0146ddbe0e6be | 6 kyu |
This Kata is based on [Number of integer partitions](https://www.codewars.com/kata/546d5028ddbcbd4b8d001254).
An *integer partition with distinct parts* of `n` is a decreasing list of positive integers which sum to `n`, in which no number occurs more than once.
For example, there are 3 integer partitions of 5:
```
[5], [4,1], [3,2].
```
~~~if-not:rust
Write a function which returns the number of integer partition with distinct parts of `n` (1 <= n <= 600).
~~~
~~~if:rust
Write a function which returns the number of integer partition with distinct parts of `n` (1 <= n <= 2821).
~~~
| algorithms | def partitions(n):
T = [1] + [0] * n
for i in range(1, n + 1):
for j in range(n, i - 1, - 1):
T[j] += T[j - i]
return T[- 1]
| Number of integer partitions with distinct parts | 6267a007e67fba0058725ad2 | [
"Mathematics",
"Algorithms",
"Discrete Mathematics"
] | https://www.codewars.com/kata/6267a007e67fba0058725ad2 | 5 kyu |
### For whom the Bell tolls
Write a function bell that will receive a positive integer and return an array. That's all splaining you receive; what needs to be done you'll have to figure out with the examples below.
```
n => resulting array
1 => [1]
2 => [2, 2]
3 => [3, 4, 3]
4 => [4, 6, 6, 4]
5 => [5, 8, 9, 8, 5]
6 => [6, 10, 12, 12, 10, 6]
7 => [7, 12, 15, 16, 15, 12, 7]
8 => [8, 14, 18, 20, 20, 18, 14, 8]
9 => [9, 16, 21, 24, 25, 24, 21, 16, 9]
10 => [10, 18, 24, 28, 30, 30, 28, 24, 18, 10]
11 => [11, 20, 27, 32, 35, 36, 35, 32, 27, 20, 11]
12 => [12, 22, 30, 36, 40, 42, 42, 40, 36, 30, 22, 12]
```
~~~if:lambdacalc
### Encodings
purity: `LetRec`
numEncoding: `BinaryScott`
export deconstructor `foldl` for your `List` encoding
~~~ | games | def bell(n):
return [(i + 1) * (n - i) for i in range(n)]
| For whom the Bell tolls | 62665d43e67fbaf7b37212d2 | [
"Algorithms",
"Logic",
"Puzzles"
] | https://www.codewars.com/kata/62665d43e67fbaf7b37212d2 | 6 kyu |
# Welcome To Pawn Promotion Kata.
In this kata you will write a program that studies a chess board
that contains only two pieces (pawn) and (king).
The pawn is always in the last row
as it will now turn into one of the four pieces
```[queen , rook , bishop , knight]``` in a process called {pawn promotion}.
Your task is to locate the king and the pawn and choose the appropriate piece
to promote it and put the king in check.
```The letter P represents the pawn```
```The letter K represents the king```
# Examples:
```
input:
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
['P', ' ', ' ', 'K', ' ', ' ', ' ', ' '],
output:
['queen', 'rook']
```
```
input:
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', 'K', ' '],
[' ', ' ', ' ', ' ', 'P', ' ', ' ', ' '],
output:
['knight']
```
```
input:
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', 'K', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '],
[' ', ' ', 'P', ' ', ' ', ' ', ' ', ' '],
output:
[]
```
# Q&A:
```
Q:- Why was the order of the pieces in the first example as ['queen', 'rook']
and not the other way around ?
```
```
A:- In the event that there is more than one option that achieves check,
return the order of the pieces as follows [queen, rook, bishop, knight]
```
# special cases:
- If there is no piece that can make check immediately upon upgrade, return [].
- If there is no pawn on the board, return [].
- If there is no king on the board, return []. | reference | def promotion(board):
pieces = [(i, j) for i, r in enumerate(board)
for j, p in enumerate(r) if p != ' ']
if not len(pieces) == 2:
return []
a, b = pieces
if a[0] == b[0] or a[1] == b[1]:
return ['queen', 'rook']
if {abs(a[0] - b[0]), abs(a[1] - b[1])} == {1, 2}:
return ['knight']
if abs(a[0] - b[0]) == abs(a[1] - b[1]):
return ['queen', 'bishop']
return []
| Pawn Promotion | 62652939385ccf0030cb537a | [
"Fundamentals"
] | https://www.codewars.com/kata/62652939385ccf0030cb537a | 6 kyu |
## Overview
An eccentric chessboard maker likes to create strange N x N chessboards.
Instead of making all the rows and the columns on his chessboards the same size, he likes to make chessboards with row and columns of varying sizes:
<svg xml:space="preserve" viewBox="0 0 500 500" height="500" width="500" version="1.1" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg">
<desc>Created with Fabric.js 4.6.0</desc>
<defs>
</defs>
<g id="pejCLSXPaosiQI-zZAy6k" transform="matrix(0.97 0 0 1.29 180 240)">
<path stroke-linecap="round" d="M -186.40124 -186.40124 L 186.40124 -186.40124 L 186.40124 186.40124 L -186.40124 186.40124 z" transform="translate(0, 0)" style="stroke: rgb(23,219,73); stroke-width: 0; stroke-dasharray: none; stroke-linecap: butt; stroke-dashoffset: 0; stroke-linejoin: miter; stroke-miterlimit: 4; fill: rgb(211,211,211); fill-rule: nonzero; opacity: 1;"></path>
</g>
<g id="AoTJfn-fmbdoEzJGz127N" transform="matrix(1 0 0 1 180 40)">
<path stroke-linecap="round" d="M -140 0 L 140 0" transform="translate(0, 0)" style="stroke: rgb(0,0,0); stroke-width: 1; stroke-dasharray: none; stroke-linecap: butt; stroke-dashoffset: 0; stroke-linejoin: miter; stroke-miterlimit: 4; fill: rgb(255,255,255); fill-opacity: 0; fill-rule: nonzero; opacity: 1;"></path>
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For example, here are the specifications of his latest 5x5 chessboard:
`cs = [ 3, 1, 2, 7, 1 ]` contains the 5 widths of the 5 columns.
`rs = [ 1, 8, 4, 5, 2 ]` contains the 5 heights of the 5 rows.
To be clear, reading the examples of `cs` and `rs` above, the width of the 4th column on the chessboard is 7, the height of the 2nd row on the chessboard is 8, etc.
Notice, by the way, that since `sum(cs) != sum(rs)`, the total width of the strange chessboard is not necessarily the same as the total height.
The chessboard maker now wants to determine how much white and black wood he needs to cover the areas in a given chessboard, so is asking you to find the **total white area** and the **total black area** on his board.
---
## Important convention
All his chessboards start with a **WHITE CELL IN THE "TOP LEFT"**, and then alternate black & white in the usual way.
So referring again to `cs` and `rs` above, the top left cell on the board is defined by the first column, of width = 3, and the first row, of height = 1, so it has an area = 3\*1 = 3 and is colored **WHITE** due to the coloring convention.
Now as we move around the board, we can consider for example the cell in the board's column 3 & row 4 - it has width = 2 and height = 5, for a total area of 2\*5 = 10. It is colored **BLACK**, as determined by the alternating chessboard coloring pattern and the starting condition of the top left cell being **WHITE**.
You may want to draw out a 5x5 board with top left cell colored white to make sure you agree with the above results.
---
## Inputs
1. You will be given two lists/arrays, `cs` and `rs`, both of size N.
2. The widths of all N columns represented in `cs`, and the heights of all N rows represented in `rs`, will always be **integers**, and >= 1
3. The size of the chessboard, i.e. the value of N, ranges in `1 <= N <= 41000`
4. **PERFORMANCE REQUIREMENTS:** given the very large N board sizes used in some of the random tests, as explained in part 3 above, **a brute force O(N^2) attempt at solving this kata will timeout...**
## Output
A tuple/struct giving `(total_white_area, total_black_area)`
---
## Example
With the values from `cs` and `rs` above:
`cs = [ 3, 1, 2, 7, 1 ]`
`rs = [ 1, 8, 4, 5, 2 ]`
There are 25 total cells on the chessboard, of which 13 are colored white and 12 are colored black.
Adding up the white cells' areas, you should find that the total_white_area is 146, and adding up the black cells' areas that the total_black_area is 134, hence the output is `(146, 134)`.
~~~if:lambdacalc
---
## Encodings
purity: `LetRec`
numEncoding: `BinaryScott`
export constructors `nil, cons` for your `List` encoding, deconstructors `fst, snd` for your `Pair` encoding.
Actual test sizes are smaller than in other languages. They will be indicated in the test headers.
~~~
---
## Note for troubleshooting
You may want to check that your solution has:
`total_area_of_board = total_white_area + total_black_area = total_width * total_height = sum(cs) * sum(rs)`
In the example above, note that `total_white_area + total_black_area = 146 + 134 = 280`, which is indeed equal to `sum(cs) * sum(rs) = 14 * 20 = 280`
| reference | def white_black_areas(cs, rs):
r_even, r_odd = sum(rs[1:: 2]), sum(rs[:: 2])
c_even, c_odd = sum(cs[1:: 2]), sum(cs[:: 2])
return (c_odd * r_odd + c_even * r_even, r_even * c_odd + r_odd * c_even)
| Find the total white and black areas in a strange chessboard | 6262f9f7afc4729d8f5bef48 | [
"Fundamentals",
"Algorithms",
"Puzzles",
"Mathematics"
] | https://www.codewars.com/kata/6262f9f7afc4729d8f5bef48 | 6 kyu |
## Task
There are several units in a line, out of your sight. You will be given a list of hints (a list or array of strings) that indicates who is next to who in the queue, and you have to rebuild the queue of people, in appropriate order.
## Example
With these hints,
```
["white has black on his left",
"red has green on his right",
"black has green on his left"]
```
you should reduce that the queue is the following:
```
["red", "green", "black", "white"]
```
### Notes:
1. It is always possible to reduce the complete order in the queue.
1. The hints can be of two different forms, such as `"white has black on his left"`, or `"black has white on his right"`.
3. Each queue has at least two units. | games | def line_up(hints):
graph = dict((a, b) if side == 'right' else (b, a)
for a, _, b, _, _, side in map(str . split, hints))
start = (set(graph) - set(graph . values())). pop()
queue = []
while start:
queue . append(start)
start = graph . get(start)
return queue
| Line up | 625d02d7a071210017c8f0c3 | [
"Arrays",
"Puzzles"
] | https://www.codewars.com/kata/625d02d7a071210017c8f0c3 | 6 kyu |
## Intro:
You may have heard in a math class somewhere that at small values of x, sin(x) can be well approximated by x (ie sin(x) looks linear near 0). This is true, and makes for some pretty slick shortcuts when calculating sin(x) at small x. You may have also heard that sin(x) can be approximated by something called a taylor sum, given an appropriate number of terms. This is also true, and the reason x is a decent approximation for sine at small x is the same reason why x is the first term in the taylor series for sine.
The taylor series of sine looks like:
```math
sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!}+...+\frac{(-1)^i}{(2i+1)!}x^{2i+1} = \sum_{i=0}^{\infty }\frac{(-1)^i}{(2i+1)!}x^{2i+1}
```
with the last term here being the generalized 'ith' term of the taylor series. When computing sums like these, generally the ith term of a series is all you need to know to start modeling a given function. Simply adding together each ith term starting at 0 and iterating on i will pull the function value closer and closer to the true value of sin(x). This series (centered at 0) is what we will be investigating in this kata.
## The Task:
In this kata, your function will be given a position (x) to evaluate the series at, and a target error bound. Your function will then calculate the minimal number of terms to approximate sin(x) within the given error bound, and return the number of terms.
Test cases here assume a taylor series centered at zero, and to avoid any ambiguity all approximations must return **at least one term.** Feel free to compare your approximation to a library's calculation of sine to determine error values, the function will not be given an error value lower than 1e-6.
An image showing the sum's accuracy improving over a few different terms:
As the degree of the Taylor polynomial rises, it approaches the correct function. This image shows sin x and its Taylor approximations by polynomials of degree 1, 3, 5, 7, 9, 11, and 13 at x = 0.
via https://en.wikipedia.org/wiki/Taylor_series | reference | from itertools import count
from math import sin
def taylors_sine_terms(x: float, epsilon: float) - > int:
sin_x = sin(x)
approx = 0
fact = 1
for i in count(1):
approx -= (- 1) * * i * x * * (2 * i - 1) / fact
if abs(sin_x - approx) < epsilon:
return i
fact *= 2 * i * (2 * i + 1)
| Taylor's Sine Approximation: How Many Terms? | 615dd90346a119004af6d916 | [
"Fundamentals"
] | https://www.codewars.com/kata/615dd90346a119004af6d916 | 6 kyu |
You will get a list with several scattered numbers
You must check that the sum of the two values on both sides equals the sum of the rest of the list elements
And if not, delete the two elements on the sides and check repeatedly,
until you reach the condition checklist:
The sum of the list without the sides = the sum of the sides
If it never equals return an empty list []
`note: list length can be up to 500 items`
# Example:
Ex1:
```python
[1,2,3,4,5] ==> 1+5 != 2+3+4 ==> [2,3,4] ==> 2+4 != 3 == [3] ==> 3+3 != 0 ==> []
```
note: (3+3) because 3 is first side and last side... (!= 0) because sum of list without sides = 0
Ex2:
```python
[0,104,3,101,0,111] ==> 0+111 != 104+3+101+0 ==> [104,3,101,0] ==> 104+0 = 3+101 ==> [104,3,101,0]
```
Ex3:
```python
[1,-1] ==> 1-1 = 0 ==> [1,-1]
```
note: (1-1) because 1 is first side and -1 is last side... (= 0) because sum of list without sides (1, -1) = 0 | reference | def plastic_balance(L):
if L and L[0] + L[- 1] != sum(L[1: - 1]):
return plastic_balance(L[1: - 1])
return L
| Plastic Balance | 625ea5c1a071210065c923af | [
"Fundamentals",
"Algorithms",
"Lists",
"Data Structures",
"Arrays"
] | https://www.codewars.com/kata/625ea5c1a071210065c923af | 7 kyu |
# Xor reduction (medium)
Given two integers, m and n, return the cumulative xor of all positive integers between them, inclusive. E.g. (0, 6), return `0^1^2^3^4^5^6` => `7`.
# Constraints:
0 <= m, n <= 10^15
m > n
# Hints
- Try the [easy variant](https://www.codewars.com/kata/56d3e702fc231fdf72001779) first
- Brute-force solutions will likely be rejected due to time-out.
- Read more about [xor](https://en.wikipedia.org/wiki/Exclusive_or).
- Look for patterns for successive n values.
| games | def xorr(n):
return (n, 1, n + 1, 0)[n % 4]
def xor_reduction(m, n):
return xorr(m - 1) ^ xorr(n)
| Xor reduction (medium) | 625d1c5b2f6c2c00300d97b7 | [
"Algorithms",
"Puzzles"
] | https://www.codewars.com/kata/625d1c5b2f6c2c00300d97b7 | 6 kyu |
Wordle is a game where you try to guess a 5-letter word. https://www.nytimes.com/games/wordle/index.html
After each guess, the color of the tiles (letters) will change to show how close your guess was to the word.
- Green (G) letter: This letter is in the correct position
- Yellow (Y) letter: This letter is present in the word, but in the wrong position
- Grey (-) letter: This letter is not in the word, except when the special rule applies (see below)
---------
For this kata, instead of playing the game like an honest person, you are to write a cheat-bot that filters down the set of words (**wordlist**) using a list of **guesses** and their respective feedbacks. For example, if you were provided this wordlist and feedbacks:
```python
{"SHEEP", "KINKY", "SWEET", "MAUVE", "FLUNG", "SKEET", ... } # wordlist
[("SPOIL", "G----"), ("STEAD", "GYG--"), ("SEETH", "GYGY-")] # guesses
```
Your output should be the set of remaining words, each satisfying all the hints in **guesses**:
```python
{'SWEET', 'SKEET'} # <- your output
```
And because you are special, this kata will not be restricted to 5-letter words!! The word length for each test case, however, would be consistent with the guesses. If the test case asks for a 9-letter word, all guesses in the hints would be of length 9. Make sure you can handle some big words!
!! There are special rules when the same character appears multiple times in the guesses and/or the solution. After assigning greens to all applicable tiles, any remaining guessed characters will be assigned yellow until no matching character is left in the solution, the rest will be assigned grey. For example, if the solution is "ATARI", and the guess is "ABBAA", the hint would be "G--Y-" | games | from collections import Counter
def wordle(wordlist, guesses):
for g, r in guesses:
z = Counter(d for d, p in zip(g, r) if p != '-')
wordlist = [w for w in wordlist if all((p == 'G') == (c == d) and not (p == '-' and w . count(
d) > z[d]) for c, d, p in zip(w, g, r)) and all(w . count(c) >= v for c, v in z . items())]
return wordlist
| Wordle! Cheat bot | 6255e6f2c53cc9001e5ef629 | [
"Puzzles"
] | https://www.codewars.com/kata/6255e6f2c53cc9001e5ef629 | 5 kyu |
__Prerequirements__
This kata is the second part of the [Barista Problem](https://www.codewars.com/kata/6167e70fc9bd9b00565ffa4e). I strongly encourage you to solve that one first.
__Context__
You got your RAISE!! YAY!!
But now, time has passed, and a position has opened up for Barista Manager.
__Description__
In order to be promoted to the rank of manager, you must prove that you have the skills required, such as managing more than one coffee machine. So your boss gives you access to a number of coffee machines that all require the same amount of time to clean. Like before, ignore the time taken to write down the orders and focus only on the time needed to clean the machines and the time needed to brew each coffee.
__Task__
Given a list of orders and a number of coffee machines, you must return the __minimum__ time needed for all of the customers to wait for their coffee, so the sum of all the waiting times.
__Note that__
* Some customers may ask you for just a glass of water, so that means that the time you would need to "brew" their "coffee" is `0`. Also, you do not need to clean anything in this case. The time you need to fill up the glass of water is also insignificant in comparison to brewing a coffee. Water is free in this cafetaria. :)
* The cleaning time for all coffee machines is `2` minutes.
* These coffee machines include a self-cleaning system, so they can all begin cleaning themselves immediately after a coffee is made.
* The time you need to walk from a coffee machine to another one, or with the coffee cup to a customer, is insignificant.
* You only have access to these machines, so you can brew up to `second parameter` coffees at the same time.
* If you receive the following orders `[2,3,4]` and you have `2` coffee machines, you could choose to brew the third and the second coffee first, so the third customer has to wait `4` minutes, and the second one has to wait `3` minutes. After that you clean the first (or the second) coffee machine, which that takes `2` minutes. Let's say that you choose to clean the first coffee machine so that means that the first customer has to wait `3` minutes for the second customer's coffee + `2` minutes for the cleaning + `2` minutes for his coffee to finish brewing. In total that would add up to `14` minutes. This is not necessarily the __minimum__ time needed.
* There may be no orders, which means that you will have to repeat the contest another day.
* The number of coffee machines will always be smaller than or equal to the number of all of the orders you receive (water glasses included) (your boss won't give you more than you can actually use, nor read the orders themselves, just count them :) ).
* There is also a person hired to clean all the coffee machines after you go home, so they are all clean the next day.
* You choose in what order you brew the coffees.
__Examples__
```
[0,0,1] , 1 => 1
[2,3,4] , 1 => 22
[2,3,4] , 2 => 13
[2,3,4] , 3 => 9
```
__Limits__
```
0 <= number of orders <= 30
0 <= size of orders <= 40
0 <= coffee machines <= number of orders
```
~~~if:lambdacalc
__Encodings__
`purity: LetRec`
`numEncoding: BinaryScott`
export constructors `nil, cons` for your `List` encoding
~~~
__Special Thanks to the great Discord community for helping with the creation of this kata and also to the programmers that helped a lot in the " discuss " section.__ | reference | def barista(coffees, n):
coffees = sorted([v for v in coffees if v])
times = coffees[: n]
for v in coffees[n:]:
times . append(times[- n] + 2 + v)
return sum(times)
| Barista Manager | 624f3171c0da4c000f4b801d | [
"Algorithms"
] | https://www.codewars.com/kata/624f3171c0da4c000f4b801d | 5 kyu |
Create a function mul37 that returns the value of num represented in the form `3 * a + 7 * b`
e.g. ```9``` would return ```3 * 3 + 7 * 0```
e.g. ```10``` would return ```3 * 1 + 7 * 1```
e.g. ```100``` would return ```3 * 3 + 7 * 13```
e.g. ```999``` would return ```3 * 4 + 7 * 141```
Note:
- a and b are integers ranging from 0 to 1000 inclusive.
- The sum of a and b is the smallest possible.
- All inputs are vaild* and 0 =< num =< 1000.
*inputs won't be in {1, 2, 4, 5, 8, 11} | reference | def mul37(num):
k = (2 * num + 6) / / 7
a, b = 7 * k - 2 * num, num - 3 * k
return f'3 * { a } + 7 * { b } '
| Numbers in terms of 3's and 7's | 62524390983b35002c8ff1e5 | [
"Puzzles",
"Mathematics",
"Fundamentals"
] | https://www.codewars.com/kata/62524390983b35002c8ff1e5 | 7 kyu |
## Steganography Part 1
Part 1 of a series using Steganography on RGB image pixels.
Part 2: https://www.codewars.com/kata/6251cba4629507b60a041b0e
---
Steganography is the practice of concealing a message within another medium.
In this case, we'll be taking a message string, and concealing it inside of RGB (Red, Green, Blue) Image Pixels using `Least Significant Bit` (LSB) *Insertion*.
Your function will take the string `msg` and build a **Bitmask** array that is then applied to the `pixels` array, returning the modified pixel values.
## Notes
1. You can assume that the `pixels` inner arrays always have a length of 3 (for Red, Green, Blue).
2. Your code should return `nil`/`None` if there aren't enough pixels to conceal the message.
- If the message has been fully concealed and there are remaining pixels, the remaining pixels should be left unmodified.
3. Since bytes are 8 bits long and `pixels` are in groups of 3, the last color of every 3rd array/ group of pixels should not be modified.
## Example
With the following inputs:
```
msg = "Hello"
pixels = [
[255, 0, 0],[0, 255, 0],[0, 0, 255],
[255, 0, 0],[0, 255, 0],[0, 0, 255],
[169, 105, 71],[172, 211, 192],[181, 140, 38],
[108, 58, 63],[105, 235, 16],[204, 69, 21],
[24, 40, 224],[88, 84, 121],[123, 41, 163],
]
```
### Bitmask Array
The `bitmask` array is a binary representation of each string character. For example with `"Hello"` the first character is `H`, and the `ASCII` value of `H` is `72`, which is is `01001000` in binary.
A `bitmask` array for `H` will look like:
```
# 01001000 -> 72 -> 'H'
bitmask = [0, 1, 0, 0, 1, 0, 0, 0]
```
Except for the last color of every third pixel,
each bit of the bitmask is applied to one color value of the pixels array, using lsb insertion (which could actually be seen as a lsb _replacement_).
Ex:
- The first color in the `pixels` array is `255` which is `11111111` in binary.
- Our first bit in the `bitmask` array is `0`, we'll need to **insert** `0` as the `LSB` of `11111111` giving us `11111110`
- _Note: you can also think of this as _replacing_ the bit that is already there with a new bit._
- This color has been modified to `254`, move on to the next color and next bit in the `bitmask` array
### Example output:
```
output = [
[254, 1, 0],[0, 255, 0],[0, 0, 255], // <- 01001000 <- 72 <- 'H'
[254, 1, 1],[0, 254, 1],[0, 1, 255], // <- 01100101 <- 101 <- 'e'
[168, 105, 71],[172, 211, 193],[180, 140, 38], // <- 01101100 <- 108 <- 'l'
[108, 59, 63],[104, 235, 17],[204, 68, 21], // <- 01101100 <- 108 <- 'l'
[24, 41, 225],[88, 85, 121],[123, 41, 163], // <- 01101111 <- 111 <- 'o'
]
```
Notice how each RGB pixel group contains off-by-one differences, this is the `bitmask` array applied to the input pixels `LSB` bit-by-bit. | reference | def conceal(msg, pixels):
bits = '' . join(f' { ord ( c ) : 0 > 8 b } .' for c in msg)
if len(bits) <= len(pixels) * 3:
bits = iter(bits)
return [[mix(v, next(bits, ".")) for v in p] for p in pixels]
def mix(v, b):
return v if b == "." else v >> 1 << 1 | int(b)
| Steganography Part 1 | 624fc787983b3500648faf11 | [
"Cryptography",
"Fundamentals"
] | https://www.codewars.com/kata/624fc787983b3500648faf11 | 6 kyu |
## Overview
In this kata you will be given:
1. A `string` of `n` inequality symbols; i.e. `>` or `<`
2. `n+1` **distinct** integers.
Your goal is to place **all** of the `n+1` integers into the available positions created by the given `n` inequality symbols, so that **the entire resulting expression is** `True`.
Just to be clear, if you have `n` inequality symbols there are `n-1` positions "inside" the string where you can place `n-1` distinct integers, and then there are `2` more positions "outside" the string - one at the beginning, and one other at the end - where you can place `2` more integers.
This is why you **will always be given `n` inequality symbols and `n+1` distinct integers**.
`_ < _ > _ > _ < _ < _ > _ < _ > _`
So, in the illustration above we have `n = 8` inequality symbols, and `n+1 = 9` empty positions, represented by `_`, in which to place the `n+1 = 9` integers.
---
## Inputs
1. `ineqs` = a `string` of `n` inequality symbols with **no spaces**, e.g. `'<><<><><'`
2. `ints` = a `tuple/vector` (depending on language) of `n+1` **distinct** integers, e.g. `(12, 9, 31, 47, 15, 11, 22, 8, 4)`
**Performance requirements:** Values up to `n = 10000` will be used in the random tests.
## Output
You will return an array **of the n+1 integers, in the order in which you want to insert them into the n+1 inequality positions**.
The tests will then insert the elements of your solution into the inequalities and check that the resulting expression evaluates to `True`.
The tests will also check that you have used **all the integers from the input**.
---
## Example, with n = 8 inequalities and n+1 = 9 integers
```python
ineqs = '<><<><><'
ints = (12, 9, 31, 47, 15, 11, 22, 8, 4)
```
For these inputs, a valid solution for you to return is - for example - the array:
```python
[8, 31, 4, 9, 47, 12, 22, 11, 15]
```
because the resulting Boolean expression:
```python
8 < 31 > 4 < 9 < 47 > 12 < 22 > 11 < 15
```
will indeed evaluate to `True`.
| reference | def interweave(a, b):
b = sorted(list(b))
i, j, s = 0, len(b) - 1, []
for c in a:
match c:
case '<':
s . append(b[i])
i += 1
case other:
s . append(b[j])
j -= 1
s . append(b[i])
return s
# from itertools import permutations, zip_longest
#
# def interweave(ineqs, ints):
# for xs in permutations(ints):
# #print(s)
# s = ''.join([str(a) if o is None else f"{a}{o}" for a,o in zip_longest(xs,ineqs)])
# r = eval(s)
# if r:
# return s
| Interweave the Inequalities and the Integers | 624dfbc87ee0d200157ad483 | [
"Fundamentals",
"Algorithms"
] | https://www.codewars.com/kata/624dfbc87ee0d200157ad483 | 6 kyu |
Although its use in screening healthy patients is debatable, Prostate Specific Antigen (PSA) is a very useful blood test for following the response of prostate cancer to treatment.
For patients treated with radiation therapy alone for prostate cancer, the PSA tends to take 2-3 years to reach a minimum value (nadir). If the PSA begins to rise again after the nadir value, this may represent a recurrence of prostate cancer.
The 1996 ASTRO (American Society for Radiation Oncology) definition (1) of a *biochemical recurrence* is:
3 consecutive rising PSA values after reaching the nadir value (lowest point) of the sample.
You are given a list of PSA values for one patient.
The list consists of 12 PSA values in ng/mL (as a float with 2 decimal places) each taken 6 months apart. The first value is taken immediately before treatment. The accuracy is two decimal points. The nadir (minimum value) will appear only once in the list.
Return ``True`` if the values meet the ASTRO criteria (3 consecutive increases in PSA occuring any time after reaching the lowest value) or ``False`` if they do not.
For clarity, the 3 consecutive increases may start directly from the nadir but do not have to. See examples below.
Examples:
```if:python
Input:
[7.91, 2.43, 1.49, 0.99, 0.74, 0.48, 0.52, 0.50, 0.66, 1.26, 1.36, 1.35]
Ouput:
True
# Nadir 0.48, 3 subsequent rises: 0.50->0.66, 0.66->1.26 , 1.26->1.36
Input:
[9.98, 8.56, 4.62, 1.16, 0.26, 0.37, 0.32, 1.02, 0.99, 1.56, 1.41, 2.35]
Ouput:
False
# Nadir 0.26. No subsequent sequence of 3 consecutive rises.
Input:
[12.57, 6.86, 1.86, 1.93, 0.60, 1.26, 0.99 ,2.1, 0.70, 0.72, 0.88, 1.9]
Output:
True
# Nadir 0.60, 3 subsequent consecutive rises: 0.70->0.72, 0.72->0.88, 0.88->1.9
Input:
[14.66, 3.14, 0.53, 0.58, 1.00, 1.26, 0.99 ,2.1, 1.50, 2.53, 2.17, 2.50]
Output:
True
# Nadir 0.53. 3 subsequent consecutive rises: 0.53->0.58, 0.58->1.00, 1.00-> 1.26
```
(1) Cox J, Grignon D, Kaplan R, Parsons J, Schellhammer P. Consensus statement: guidelines for PSA following radiation therapy. Int J Radiat Oncol Biol Phys. 1997;37:1035– 1041.
| reference | def recurrence(values):
# let's take the first item as nadir, the previous item is temporarily equal to nadir
prev = nadir = values[0]
counter = 0 # a counter of ever higher successive values
bio = False # 3 consecutive rising PSA values after reaching the nadir value, now it's False
for val in values[1:]: # get next value from the list
if val < nadir: # new nadir?
nadir = val # if yes we start counting again, so:
bio = False # - reset the bio
counter = 0 # - reset the counter
elif val > prev: # if we detected an increase:
counter += 1 # - mark it
else: # if we detected not an increase:
counter = 0 # - reset the counter to count again
prev = val # remember previous item
if counter > 2: # if we have 3 consecutive rising PSA values:
bio = True # - we found 'biochemical recurrence'
return bio
| Prostate cancer recurrence rates (ASTRO) | 624e0a4c3e1d7b0031588666 | [
"Arrays",
"Fundamentals"
] | https://www.codewars.com/kata/624e0a4c3e1d7b0031588666 | 7 kyu |
# Summary:
[Rock, Paper, Scissors](https://en.wikipedia.org/wiki/Rock_paper_scissors) is too boring- let's spice it up a little!
Given an array of possible choices, what player 1 chooses and what player 2 chooses, create a function which determines who wins.
If player 1 wins, return 'Player 1 won!'
If player 2 wins, return 'Player 2 won!'
If the game is drawn, return 'Draw!'
```if:c
( Note that in the C translation,
return values will be taken from an enum:
`enum infinite {DRAW, ONE, TWO};` )
```
Player 1 and 2 will always choose an item from the array choices.
The array will only have one of each choice.
# Now for the details:
The array will look something like this 👇
```
["rock","paper","scissors"]
```
The rules of rock-paper-scissors are as follows:
It is a tie if player 1 and player 2 choose the same option.
If player 1 chooses "paper" and player 2 chooses "rock", since paper beats rock, return 'Player 1 won!'
If player 1 chooses "paper" and player 2 chooses "scissors", since scissors beats paper, return 'Player 2 won!'
If player 1 and player 2 both choose "scissors", return 'Draw!'
(Think about how the order of each item in the array dictates if it wins or loses to other items, depending on their order)
What if the array was this? 👇
```
["scissors","paper","rock"]
```
In this case, the rules would be inverted:
Rock beats paper
Paper beats scissors
Scissors beats rock
The order of who-beats-what will change depending in their order in the array.
We don't need to limit ourselves to only rock, paper and scissors though.
Now what if the array looks like this? 👇
```
["rock","spock","paper","lizard","scissors"]
```
Here, if player 1 chooses "rock" and player 2 chooses either "spock" or "paper", return 'Player 2 won!'
If instead player 1 chooses "rock" and player 2 chooses either "lizard" or "scissors", return 'Player 1 won!'
If player 1 chooses "lizard" and player 2 chooses "paper", return 'Player 1 won!'
What if the array's length was even? 👇
```
["heavy","sniper","soldier","pyro"]
```
Here, if player 1 chooses "heavy", then player 1 wins if player 2 chooses "pyro", and loses if player 2 chooses "sniper".
However, if player 1 chooses "heavy" and player 2 chooses "soldier", then it is a draw.
Similarly, if player 1 chooses "pyro" and player 2 chooses "sniper", it is a draw.
# Edge cases:
If the array's length is 1, it will always be a draw. The same happens if the array's length is 2. | reference | def winner(choices, p1, p2):
if p1 == p2:
return "Draw!"
r = (choices . index(p2) - choices . index(p1)) % len(choices)
if r == len(choices) / 2:
return "Draw!"
if r < len(choices) / 2:
return "Player 2 won!"
return "Player 1 won!"
| Rock paper scissors infinite | 62443a1ea8fca9002346d72c | [
"Games",
"Puzzles",
"Fundamentals"
] | https://www.codewars.com/kata/62443a1ea8fca9002346d72c | 6 kyu |
The knight is a piece in the game of chess. It may move two squares vertically and one square horizontally or two squares horizontally and one square vertically. For a visual representation of the possible moves, take a look [here](https://en.wikipedia.org/wiki/Knight_(chess)#Movement) (second image).
Given the current position of the knight on the chessboard (e.g.`"b8"`) return all its possible next positions, alphabetically sorted (e.g. `["a6", "c6", "d7"]`)
```if:python,ruby
Your code can be maximum ~~170~~ 150 characters long (I underestimated your skills ;-)
```
<br>
**Note:** this is the code golf version of [this kata](https://www.codewars.com/kata/5b5736abf1d553f844000050)
---
### My other katas
If you enjoyed this kata then please try [my other katas](https://www.codewars.com/users/anter69/authored)! :-)
#### *Translations are welcome!*
...except for Ruby, which will arrive soon | games | def next_pos(s): return [f + r for f in 'abcdefgh' for r in '12345678' if ~ int(f + r + s, 24) * * 4 % 577 ^ 4 < 4]
| [Code Golf] Knight's Next Position | 6243819a58ad06b6c663d32b | [
"Restricted",
"Puzzles"
] | https://www.codewars.com/kata/6243819a58ad06b6c663d32b | 6 kyu |
# Superheroes Convention #1 Pandemic
You are organizing the next comic book convention in the famous San Diego Convention Center. Because of a new virus the convention will be just a 2-day event.
Each superhero will have his stand but there is a <strong><u>big problem:</u> </strong>They dont want to share the space with superheros they hate. For example, if Batman hates Superman, you can't put them on the same day. You may put Batman stand on day 1 and Superman stand on day 2.
Based on that, your team has given you a database for all superheroes assisting which superheroes they hate. Your job is to tell your team if it is possible to organize all the superheroes in these 2 days without sharing the place with heroes they don't like.<br/>
Be aware that hundreds of superheros attend the convention, so you should optimize your algorithm!<br/>
<strong>Assumptions:</strong> For simplicity (but not necessity/sufficiency) <br/>
<ul>
<li>Suppose the hate is mutual, so if Spiderman hates Captain America then Captain America hates spiderman.</li>
<li>Assume the superheroes dont have depression, i.e. they dont hate themselves.</li>
</ul>
<strong>Input:</strong>(dictionary) You will be receiving a Dictionary, each key is a superhero ID and its value is a list of superheroes IDs he/she hates. If the superhero doesn't hate anyone, the value is an empty list.
<strong>Output:</strong>(boolean) True if it's possible to organize all the superheroes in these 2 days without sharing the space with superheroes they hate, otherwise False.<br/>
### Example:
<strong>Input: </strong>
{0:[1, 2],
1: [0],
2: [0],
3: []}<br/>
Dont get confused, the above superhero IDs dictionary is equivalent to:<br/>
{"Superman":["Batman", "Aquaman"],
"Batman": ["Superman"],
"Aquaman": ["Superman"],
"Wonderwoman": []}<br/>
<strong>Output: </strong>True.<br>
One combination of organizing them is: Batman and Aquaman on day 1 and Superman on day 2. Wonderwoman has no conflicts so she could go any day
```python
is_possible({0:[1, 2], 1:[0], 2:[0], 3:[]}) = True
```
```javascript
isPossible({0:[1, 2], 1:[0], 2:[0], 3:[]}) = true
```
<strong>Hint:</strong> Try to sketch this using graph theory. | algorithms | def is_possible(db: dict) - > bool:
db = dict(db)
stk = list(db)
day = [0] * len(db)
while stk:
a = stk . pop()
if a not in db:
continue
lst = db . pop(a)
day[a] = day[a] or 1
oppDay = day[a] ^ 3
for b in lst:
if day[b] and day[b] != oppDay:
return False
day[b] = oppDay
stk . extend(lst)
return True
| Superheroes Convention #1 Pandemic | 6202149e89771200306428f0 | [
"Algorithms",
"Puzzles",
"Performance",
"Graph Theory"
] | https://www.codewars.com/kata/6202149e89771200306428f0 | 4 kyu |
# Elevator malfunctioning
The elevator in my building of `n` floors is malfunctioning.
Whenever someone wants to go up, the elevator moves up by `x` floors if it can. If the elevator cannot move up by `x` floors, it stays in te same spot. On the other hand, whenever someone wants to go down, the elevator moves down by `y` floors if it can. If it cannot, then the elevator stays in the same spot.<br/>
For example, in a 10-story building, the down button is programmed to move the elevator down 5 floors, while the up button moves it up 3 floors. If you are on the 4th floor and press the down button, the elevator will remain on the 4th floor. Similarly, if you press the up button from the 8th floor, the elevator will stay on the 8th floor.
## Task
The mission to accomplish is to know the minimum stops the elevator has to make(starting in the 1st floor) to reach my living floor (if possible).
## Input
up: (int) how many floors the elevator moves up.<br/>
down: (int) how many floors the elevator moves down.<br/>
n_floors: (int) the number of floors the building has. `$1\leq$`n_floors`$\leq 10^6$`<br/>
floor:(int) The floor where you live.<br/>
## Output
(int) The minimum number of stops that it takes to reach the floor where you live, otherwise -1.
<strong>Example</strong><br/>
In a 25-story building numbered from 1 to 25, the up button is programmed to take the elevator up 8 floors while the down button takes it down 11 floors. If you live on the 14th floor, the minimum number of stops required to reach your floor is 4. One optimal floor sequence** for reaching the 14th floor is (up, up, up, down): First go up to the 9th floor, then up to the 17th floor, up to the 25th floor, and finally down to the 14th floor, requiring a total of 4 stops.<br/>
```python
get_stops(8, 11, 25, 14) = 4
```
```javascript
getStops(8, 11, 25, 14) = 4
```
```java
Elevator.getStops(8, 11, 25, 14) = 4
```
**This is not always an unique sequence, there are other floor sequences that could give the minimum number of stops. In this example, other floor sequence is (up, up, down, up): Going up to 9th floor, then up to the 17th floor, down to the 6th floor and up to the 14th floor give us the same result (4 stops) but with different floor sequence. <br/>
### Notes
<ul>
<li>This problem was inspired from Puzzle 9 section 1,<i>Math Puzzles Volume 1</i>, Presh Talwalkar and from Augsburgs College math newsletter (volume 17, number 9, February 4, 2004)<a href="https://web.augsburg.edu/mathematics/laugarithms/Augarithms_v17.9.pdf">link</a></li>
</ul>
| algorithms | # return the minimum number of stops, otherwise -1
def get_stops(up, down, n_floors, floor):
visited = {1}
queue = [(1, 0)]
while queue:
fl, stops = queue.pop(0)
if fl == floor: return stops
for f in [fl+up, fl-down]:
if 1 <= f <= n_floors and not f in visited:
visited.add(f)
queue.append((f, stops+1))
return -1 | Elevator malfunctioning | 62396d476f40250024bcfce9 | [
"Data Structures",
"Algorithms",
"Mathematics",
"Puzzles"
] | https://www.codewars.com/kata/62396d476f40250024bcfce9 | 5 kyu |
The 2020 International Society of Hypertension Practice Guidelines describe raised blood pressure (hypertension) and its complications as the leading cause of death in the world at the time of publication. Blood pressure is the measure of the heart's pumping pressure against the blood vessel walls. It is recorded as a systolic pressure over a diastolic pressure (for example '120/80 mmHg') which represent the maximum and minimum pressure in the vessels between heartbeats.
For the purpose of this kata, hypertension can be diagnosed when a patient's blood pressure meets any of the three criteria below:
+ Average of all measured systolic pressures >= 140 mmHg. Must have minumum of 2 measurements.
+ Average of all measured diastolic presures >= 90 mmHg. Must have minimum of 2 measurements.
+ Any one measurement with systolic pressure >= 180 mmHg AND diastolic pressure >= 110 mmHg
You are given a list, ***patients***, in which each element represents a patient. Each patient/element consists of a list of strings (representing all measurements of a patient's blood pressure) in the format "s/d", where s is the systolic pressure and d the diastolic, both in mmHg. Systolic and diastolic numbers will always be whole numbers.
Return the number of patients that meet any of the above hypertension definitions.
Example:
```if:python,javascript
Input:
[
["130/90","140/94"],
["160/110"],
["200/120"],
["150/94","140/90","120/90"],
["140/94","140/90","120/80","130/84"]
]
Output:
3
# patients at index 0, 2, and 3 are hypertensive
```
```if:lua
Input:
{
{"130/90","140/94"},
{"160/110"},
{"200/120"},
{"150/94","140/90","120/90"},
{"140/94","140/90","120/80","130/84"}
}
Output:
3
-- patients at index 1, 3, and 4 are hypertensive
```
Constraints:
0 < length of patients list <= 10000
0 < length of element within patients <= 30 | reference | def c1(p): return (l: = len(p)) > 1 and sum(int(e . split('/')[0]) for e in p) / len(p) >= 140
def c2(p): return (l: = len(p)) > 1 and sum(int(e . split('/')[1]) for e in p) / len(p) >= 90
def c3(p): return any(
int(e . split('/')[0]) >= 180 and int(e . split('/')[1]) >= 110 for e in p)
def hypertensive(P): return sum(c1(p) or c2(p) or c3(p) for p in P)
| Diagnosing Hypertension | 621470ede0bb220022c9e396 | [
"Strings",
"Arrays",
"Fundamentals"
] | https://www.codewars.com/kata/621470ede0bb220022c9e396 | 6 kyu |
If you're not familiar with the fantastic culinary delights of the British Isles you may not know about the bread sandwich.
The idea is very simple - if you have a slice of bread between two other slices of bread, then it's a bread sandwich. Additionally, if you have a bread sandwich between two other slices of bread, you get a bread sandwich sandwich, and so on.
In this kata, we will define the following terms like so:
- *Sandwich* - Two slices of bread which may or may not have a filling
- *Bread Sandwich* - Two slices of bread which contains one slice of bread in the middle as a filling
You will need to build two functions:
- Given the number of slices of bread, return the phrase used to refer to the sandwich
```
2 - 'sandwich'
5 - 'bread sandwich sandwich'
```
- The reverse function - given the name of the sandwich, return how many slices of bread there are in the sandwich
```
'bread sandwich' - 3
'sandwich sandwich' - 4
```
- You should return `None`/`null` if there is no input / the input is invalid / there is no sandwich
- Maximum 100 slices of bread | algorithms | def slices_to_name(n):
if isinstance(n, int) and n > 1:
return ' ' . join(n % 2 * ['bread'] + n / / 2 * ['sandwich'])
def name_to_slices(name):
if isinstance(name, str):
n = 2 * name . count('sandwich') + name . count('bread')
if name == slices_to_name(n):
return n
| bread sandwiches | 622a6a822494ab004b2c68d2 | [
"Strings",
"Algorithms"
] | https://www.codewars.com/kata/622a6a822494ab004b2c68d2 | 6 kyu |
Your task here is to write [unit tests](https://en.wikipedia.org/wiki/Unit_testing) for several statements in the following `PhoneBook` class, which manages each user's phone number:
~~~if:python
```python
from collections import UserDict
class PhoneBook(UserDict):
def create(self, name: str, phone_number: str) -> None:
self[name] = phone_number # Statement 1
def retrieve(self, name: str) -> str:
if name not in self:
raise KeyError('Name not found!') # Statement 3
return self[name] # Statement 2
def update(self, name: str, phone_number: str) -> None:
if name not in self:
raise KeyError('Name not found!')
self[name] = phone_number # Statement 4
def delete(self, name: str) -> None:
if name not in self:
raise KeyError('Name not found!')
del self[name] # Statement 5
```
~~~
As you can see, there are 5 statements that are marked with `# Statement N`. Your goal is to write tests for them.
Some details about how your code is tested:
1. There are two stages of tests -
- In the first stage, your code will be used to test the `PhoneBook` class you just saw. To pass this stage, your code should <span style="color:red">cover all 5 above mentioned statements</span>.<br>
- In the second stage, your code will be used to test several incorrectly implemented `PhoneBook` classes. To pass this stage, you have to <span style="color:red">make the correct assertions</span>. In other words, your code should be able to tell us that these `PhoneBook` classes are not correct.
2. You only need to write tests for statements that are marked with `# Statement N`.
3. Do not rename existing test methods.
Note:
1. The `PhoneBook` class is already imported for you.
~~~if:python
2. If you are not familiar with writing unit tests, [this](https://docs.python.org/3.11/library/unittest.html) can be helpful.
~~~ | reference | from unittest import TestCase
class TestPhoneBook (TestCase):
def setUp(self):
# Arrange
self . phone_book = PhoneBook()
def test_create_method(self):
# Act
self . phone_book . create('User', '12345678')
# Assert
self . assertEqual(self . phone_book['User'], '12345678')
def test_retrieve_method_when_the_name_exists(self):
# Arrange
self . phone_book['User'] = '12345678'
# Act
result = self . phone_book . retrieve('User')
# Assert
self . assertEqual(result, '12345678')
def test_retrieve_method_when_the_name_does_not_exist(self):
# Act & Assert
with self . assertRaises(KeyError) as cm:
self . phone_book . retrieve('User')
# Assert
self . assertEqual(str(cm . exception), "'Name not found!'")
def test_update_method_when_the_name_exists(self):
# Arrange
self . phone_book['User'] = '12345678'
# Act
self . phone_book . update('User', '87654321')
# Assert
self . assertEqual(self . phone_book['User'], '87654321')
def test_delete_method_when_the_name_exists(self):
# Arrange
self . phone_book['User'] = '12345678'
# Act
self . phone_book . delete('User')
# Assert
self . assertNotIn('User', self . phone_book)
| Basic Unit Test Practice | 620bce448e757800242821fe | [
"Fundamentals",
"Tutorials"
] | https://www.codewars.com/kata/620bce448e757800242821fe | 6 kyu |
Popular gambling game **Joker** consists of drawing 6 balls from a drum on which are written numbers
between 1 and 45. The winning joker number is formed from the drawn numbers so that in the same order in which the balls are drawn, we write down their last digits.
For example, if the numbers 12, 35, 1, 2, 23 and 39 are drawn in order, then it is a joker No. 251239.
```1[2], 3[5], [1], [2], 2[3], 3[9] --> 251239```
Players buy winning tickets and look forward to drawing and calculating the joker number to calculate their possible gain. The winnings of the ticket are calculated by comparing the serial number written on the ticket with the calculated joker number in a way to count how many *last digits* of these two numbers match, as shown in the following table.
```
Serial No. | Name of prize
-----------+---------------
251239 | I type
X51239 | II type
XX1239 | III type
XXX239 | IV type
XXXX39 | V type
XXXXX9 | Losing card
XXXXXX | Losing card
```
In the left column are the serial numbers of the tickets, where X denotes arbitrary digits. In the right column of the table are the winnings names for each of the serial number forms. So if the serial number is exactly equal to the Joker number then we get the gain of type I, if the last 5 digits are equal we get the gain of type II, and so on, until the gain of type V, the kind we get when the last two digits are equal. All other serials are not winning.
Write a program that will return the names of the winnings for a given Joker number and given tickets. The names of the winnings must be written exactly as written in the table above (the number of winnings is written in Roman numbers with capital letters ‘I’ and ‘V’, followed by a space and lowercase letters ‘type’).
## Input
joker_nums --> an array of 6 numbers from which you have to make the winning Joker number
ticket_serials --> an array of 3 ticket serial numbers
## Output
You should return a list with the names of the winnings for the 3 given tickets
```python
joker_card([12, 35, 1, 2, 23, 39], ['151239', '251229', '251339']) --> ["II type", "Losing card", "V type"]
```
## Note
- some tickets may have leading zeros
- some joker numbers may have leading zeros
- leading zeros still count towards the total
Problem taken from the [Cantonal competition in Informatics 2016, Tuzla](https://upinitk.ba/wp-content/uploads/2018/01/Bilten2016.pdf) | reference | def joker_card(joker_nums, ticket_serials):
prize_dict = {
0: 'Losing card',
1: 'Losing card',
2: 'V type',
3: 'IV type',
4: 'III type',
5: 'II type',
6: 'I type'
}
prize = []
joker = list(map(lambda x: str(x)[- 1], joker_nums))[:: - 1]
for tricket in ticket_serials:
count = 0
for i, num in enumerate(tricket[:: - 1]):
if num == joker[i]:
count += 1
else:
break
prize . append(prize_dict[count])
return prize
| Gambling Game Joker | 621e323a98afab001628d9a0 | [
"Fundamentals",
"Strings"
] | https://www.codewars.com/kata/621e323a98afab001628d9a0 | 6 kyu |
Every book has `n` pages with page numbers `1` to `n`. The `summary` is made by adding up the **number of digits** of all page numbers.
Task: Given the `summary`, find the number of pages `n` the book has.
### Example
If the input is `summary=25`, then the output must be `n=17`: The numbers 1 to 17 have 25 digits in total: `1234567891011121314151617`.
Be aware that you'll get enormous books having up to 100.000 pages.
All inputs will be valid.
Here's the opposite kata, ***which is more complex** ([Paginating a huge book](https://www.codewars.com/kata/55905b7597175ffc1a00005a/python))*
| algorithms | def amount_of_pages(summary):
# 1-9: 9 = 9 * 1 * 10**0
# 10-99: 180 = 9 * 2 * 10**1
# 100-999: 2700 = 9 * 3 * 10**2
# 1000-9999: 36000 = 9 * 4 * 10**3
# 10000-99999: 450000 = 9 * 5 * 10**4
res = 0
for x in range(1, 6):
y = 9 * x * 10 * * (x - 1)
if summary <= y:
return res + summary / / x
res += y / / x
summary -= y
| How many pages in a book? | 622de76d28bf330057cd6af8 | [
"Puzzles",
"Algorithms"
] | https://www.codewars.com/kata/622de76d28bf330057cd6af8 | 6 kyu |
This kata is the performance version of [Don't give me five](https://www.codewars.com/kata/5813d19765d81c592200001a) by [user5036852](https://www.codewars.com/users/user5036852).
Your mission, should you accept it, is to return the count of all integers in a given range which *do not* contain the digit 5 (in base 10 representation). <br>
You are given the start and the end of the integer range. The start and the end are both inclusive.
**Examples:**
```
1,9 -> 1,2,3,4,6,7,8,9 -> return 8
4,17 -> 4,6,7,8,9,10,11,12,13,14,16,17 -> return 12
```
The result may contain fives. ;-)<br>
The start will always be smaller than the end. Both numbers can be also negative.
The regions can be very large and there will be a large number of test cases. So brute force solutions will certainly time out!
Good luck, warrior! | algorithms | import re
def dont_give_me_five(start, end):
def count(n): return int(re . sub(r'5(\d*)', lambda m: '4' + '9' *
len(m[1]), str(n)). translate(str . maketrans("56789", "45678")), 9)
if start > 0:
return count(end) - count(start - 1)
elif end < 0:
return count(- start) - count(- end - 1)
else:
return count(end) + count(- start) + 1
| Don't give me five! Really! | 621f89cc94d4e3001bb99ef4 | [
"Algorithms",
"Mathematics",
"Performance"
] | https://www.codewars.com/kata/621f89cc94d4e3001bb99ef4 | 4 kyu |
Alice and Bob are playing a game. There are `$n (1 \leq n \leq 10^{18}) $` cards on the table. The aim of the game is to collect the most cards. The rules:
* if the number of cards is *even*, the players can either take half of the cards from the stack, or only 1 card -- as they choose;
* if the number of cards is *odd*, the players must take 1 card.
Alice starts the game.
Return the maximum number of cards Alice can collect, if Bob plays optimally (tries to get as many cards as possible for himself).
## Example
When there are 10 cards, Alice can collect maximum 8, like this:
```
1. Alice takes half of the 10 cards (now has: 5) --> 5 cards left
2. Bob must take 1 card (now has: 1) --> 4 cards left
3. Alice takes half of the 4 cards (now has: 5 + 2 = 7) --> 2 cards left
4. Bob takes 1 (or half of the 2 cards) (now has: 1 + 1 = 2) --> 1 card left
5. Alice takes the last card (now has: 7 + 1 = 8) --> game over!
Thus, Alice has 8 cards and Bob has 2
``` | algorithms | def get_card(n):
if (n % 2 == 0 and (n / / 2) % 2 == 1) or n == 4:
return n / / 2
else:
return 1
def card_game(n):
alice = 0
bob = 0
flag = True
while n > 0:
this_turn = get_card(n)
if flag:
alice += this_turn
else:
bob += this_turn
flag = not flag
n -= this_turn
return alice
| Card Game | 61fef3a2d8fa98021d38c4e5 | [
"Mathematics",
"Algorithms",
"Performance"
] | https://www.codewars.com/kata/61fef3a2d8fa98021d38c4e5 | 5 kyu |
## Objective
Create a [rasterized](https://en.wikipedia.org/wiki/Rasterisation) binary image of a triangle.
### Given
- `(x1, y1)`, `(x2, y2)`, and `(x3, y3)` - **Integer** coordinates of triangle vertices.
- `n` - Size of the output image.
### Expected Output
- An `n`-by-`n` image of `1`s and `0`s.
### Details
- `1 <= n <= 100`
- The origin of the coordinate system is the upper-left pixel with coordinates `(0,0)`.
- The `x` coordinate increases to the right and the `y` coordinate increases downward.
- A pixel should be set to `1` only if it's coordinate is either inside the triangle **or on it's boundary.**
- Triangle coordinates may be outside the image bounds.
- No degenerate triangles will be provided.
## Example
<p align="center">
<img src="https://i.imgur.com/JJz4s0l.png" />
</p>
The triangle shown above is defined by vertices `(2,1)`, `(0,3)`, and `(5,4)`.
For `n=6` the expected output is
```python
[ [0,0,0,0,0,0],
[0,0,1,0,0,0],
[0,1,1,1,0,0],
[1,1,1,1,1,0],
[0,0,0,0,0,1],
[0,0,0,0,0,0] ]
```
For `n=3` the expected output is
```python
[ [0,0,0],
[0,0,1],
[0,1,1] ]
```
| algorithms | def half_plane(p1, p2, p3):
"""
Returns a function that checks the affiliation of a point (x, y)
to a half-plane (defined by point p1, p2) to which point p3 belongs.
If the given three points do not form a triangle (only a straight line),
an exception is raised.
Note that the function works for any lines also for vertical lines.
"""
(x1, y1), (x2, y2), (x3, y3) = p1, p2, p3
left, right = (y3 - y2) * (x2 - x1), (x3 - x2) * (y2 - y1)
if left > right:
return lambda x, y: (y - y2) * (x2 - x1) >= (x - x2) * (y2 - y1)
elif left < right:
return lambda x, y: (y - y2) * (x2 - x1) <= (x - x2) * (y2 - y1)
raise ValueError("Detection of a degenerate triangle.")
def draw_triangle(triangle, n):
p1, p2, p3 = triangle
p1p2 = half_plane(p1, p2, p3)
p2p3 = half_plane(p2, p3, p1)
p1p3 = half_plane(p1, p3, p2)
# The point (x, y) lies in the triangle if all three dependencies are satisfied:
return [[p1p2(x, y) and p2p3(x, y) and p1p3(x, y) for x in range(n)] for y in range(n)]
| Rasterize a Triangle | 621ab012ed37430016df15c0 | [
"Geometry",
"Graphics",
"Image Processing",
"Algorithms"
] | https://www.codewars.com/kata/621ab012ed37430016df15c0 | 6 kyu |
## Task
In this task, you need to restore a string from a list of its copies.
You will receive an array of strings. All of them are supposed to be the same as the original but, unfortunately, they were corrupted which means some of the characters were replaced with asterisks (`"*"`).
You have to restore the original string based on non-corrupted information you have. If in some cases it is not possible to determine what the original character was, use `"#"` character as a special marker for that.
If the array is empty, then return an empty string.
### Examples:
```
input = [
"a*cde",
"*bcde",
"abc*e"
]
result = "abcde"
input = [
"a*c**",
"**cd*",
"a*cd*"
]
result = "a#cd#"
``` | reference | def assemble(input):
result = list(input[0]) if input else []
for i in input:
for j, k in enumerate(i):
result[j] = k if result[j] == '*' else result[j]
return '' . join(result). replace('*', '#')
| Assemble string | 6210fb7aabf047000f3a3ad6 | [
"Strings",
"Arrays",
"Fundamentals"
] | https://www.codewars.com/kata/6210fb7aabf047000f3a3ad6 | 6 kyu |
It is hypothesized that the self-replicating RNA molecules played an early and central role in the origin of life. For more information, see [The RNA World](https://en.wikipedia.org/wiki/RNA_world).
Through phosphodiester bondings, four kinds of nucleoside triphosphate came together to form a chain that is moderately stable. These nucleoside triphosphates are
<pre>
Adenosine Triphosphate, ATP (A)
Uridine Triphosphate, UTP (U)
Cytidine Triphosphate, CTP (C)
Guanosine Triphosphate, GTP (G)
</pre>
However, the chain formed has limited length and breaks easily. 😞 Life has to wait!
But miracles do happen! When AUCG molecules form a chain long enough without breaking, they can make a loop and pair with themselves again (through intra-molecular hydrogen bonding). This kind of secondary structure is called Stem-loop/Hairpin-loop where the stem part is super stable.<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/3/3f/Stem-loop.svg/1024px-Stem-loop.svg.png"/>Life always finds a way!😎</br>
You will be given a single-stranded RNA floating on the oceans of proto-earth, in string format like 'AGCUGG'. It could be 4 to 500 char long. Help the strand make its longest stem possible and <u>return the length of the stem in terms of no. of base pairs.</u></br>
The unit for length of the stem will be in <u>base pairs</u>.
<pre>
Given that:</br>
1. The base pairs are of two types: A-U and G-C
2. Minimum length for a stem to form is 4 base pairs
3. You are not allowed to form more than 1 loop (to avoid complexity)
4. The smallest loop has 6 unbonded bases (unbonded as in absence of base pairing)
5. Loop with more than 12 unbonded bases is extremely unstable (12 is the limit)
</pre>
<u>Example</u>: 'AAUUUGAUCUCAAUUU'
<pre>
G
U A
A A U U
| | | | U
U U U A A
C C
U
<u>should give:</u> 4 as output</br>
</pre>
If no stem formed, return 0 | algorithms | def longest_stem_length(rna):
min_stem, loop_sizes, base_pairs = 4, range(
6, 13), dict(zip('ACGU', 'UGCA'))
lgt, pairs = 0, [(i, i + s + 1) for s in loop_sizes for i in range(
len(rna) - s - 1) if rna[i + s + 1] == base_pairs[rna[i]]]
while pairs:
lgt, pairs = lgt + 1, [(i - 1, j + 1) for i, j in pairs if 0 <= i -
1 and j + 1 < len(rna) and rna[j + 1] == base_pairs[rna[i - 1]]]
return 0 if lgt < min_stem else lgt
| 🧬Cooking life on Proto Earth🌎 | 5fe2ea999fd2140016feec63 | [
"Strings",
"Algorithms"
] | https://www.codewars.com/kata/5fe2ea999fd2140016feec63 | 5 kyu |
🚩 What is the flag?
- The flag is hidden in the string variable `FLAG`
- The flag is in the format `FLAG{CAPITAL-LETTERS-WITH-HYPHEN}`
- The flag couldn't be found from the string literal right away... but *You can indeed **see** it somehow!*
- A magic number `16` might help... but how should it be used?
| games | # WTF?
flag = "FLAG{IS-IT-EASY-YET-FUN}"
| What is the flag? | 61efc02e4fd88600343b5c58 | [
"Fundamentals",
"Puzzles",
"Strings"
] | https://www.codewars.com/kata/61efc02e4fd88600343b5c58 | 7 kyu |
### Overview
We often use smiley faces in correspondence with other people. They allow us to quickly show our reaction to the person(s) we are talking to.
But one day you wanted to make your correspondence more joyful. So today you have the opportunity to make it happen.
### Task:
In this kata, your task will be to replace sad emoticons with funny ones.
The emoticons, will be represented from:
* Eyes: marked as `:`, `;` or `=`
* Nose (optional): marked as `-` or `~`
* Mouth: marked as `(` or `[`
#### Examples:
```javascript
smile("Howdy :(") // should return "Howdy :)"
smile("Never be sad =[") // should return "Never be sad =]"
smile("Change this `=(` and this `:[`") // should return "Change this `=)` and this `:]`"
``` | algorithms | def smile(text):
eyes = [":", ";", "="]
noses = ["", "-", "~"]
mouths = ["[", "("]
mouths_2 = [']', ')']
for eye in eyes:
for nose in noses:
for i, mouth in enumerate(mouths):
face = eye + nose + mouth
face_2 = eye + nose + mouths_2[i]
if face in text:
text = text . replace(face, face_2)
return text
| Make everyone happy :) | 61eeb6e7577f050037b17a2d | [
"Fundamentals",
"Algorithms",
"Strings"
] | https://www.codewars.com/kata/61eeb6e7577f050037b17a2d | 6 kyu |
The Royal Household is opening a new vacancy of microwave oven maid, who will heat up dishes for Queen Elizabeth II. All candidates must pass several very difficult tests and training in boot camp. Good luck, recruit!
### The touchpad test.
On a microwave the maximum number that you can enter for minutes or seconds is `99`. Hence, the time can be represented in two ways:
1. using the standard method where seconds go up to `59` only,
1. using an alternative method where seconds can go up to `99` just like minutes do.
A good microwave maid can't waste her time moving finger unnecessarily from one button to another to set the timer.
Given the time required to heat up the food in seconds, you must return a string of digits that the maid has to input using one of the aforementioned methods such that the number of different consecutive digits is minimal. If the number of differences is the same, pick the shorter string. If both string have the same length too, use the standard method.
#### Example 1:
```
input = 71
method 1: "111" (1 minute and 11 seconds)
method 2: "71" (0 minutes and 71 seconds)
return "111" since it has fewer different consecutive digits
```
#### Example 2:
```
input = 72
method 1: "112" (1 minute and 12 seconds)
method 2: "72" (0 minutes and 72 seconds)
return "72" since the number of different consecutive digits is the same but "72" has fewer digits than "112"
```
#### Example 3:
```
input = 3690
method 1: "6130" (61 minutes and 30 seconds)
method 2: "6090" (60 minutes and 90 seconds)
return "6130" (the standard way) since both the number of different consecutive digits and the number of digits are the same
```
If you like this kata, check out the another one: [Maid Ep2: Christmas Trash Party](https://www.codewars.com/kata/639df18a29d4991916ba5cdd)
| reference | import re
def toS(m, s): return f' { m }{ s : 0 > 2 } ' . lstrip('0')
def compact(s): return re . sub(r'(.)\1+', r'\1', s)
def cmpS(a, b): return (len(a) > len(b)) - (len(a) < len(b))
def get_best_combination(t):
m, s = divmod(t, 60)
M, S = m - 1, s + 60
a, b = map(toS, (m, M), (s, S))
A, B = map(compact, (a, b))
cmp = cmpS(A, B) or cmpS(a, b)
return a if s > 39 or cmp < 1 else b
| Microwave maid Ep1: The touchpad test | 61e1f175fbf3bd002a5528cd | [
"Fundamentals",
"Strings"
] | https://www.codewars.com/kata/61e1f175fbf3bd002a5528cd | 6 kyu |
Let `$f(s) = s_1 \cdot 1 + s_2 \cdot 2 + ... + s_N \cdot N$` where `$N \leq 2 \cdot 10^4$` is the length of an array `$s$`.
In other words, `$f(s) = \sum_{i=1}^{N} s_i \cdot i$`.
Given a sequence of integers, `$a$` (`$-10^9 \leq a_i\leq 10^9$`) and an array of queries, `$qs$`, where each `$qs_i=[l,r]$` (`$l \leq r$`) describes the indexes for that query (inclusive), find the value of `$f([a_{l}, a_{l+1}, ..., a_{r-1}, a_{r}])$` for each query and returns the result as an array.
Notes:
* The sequences here are 1-indexed, meaning `$s_1$` is the first element of the sequence `$s$`.
* Watch out for performances: anything quadratic or worse will time out.
* 3 batches of random tests (small, medium, performance)
```if:python
* The last batch of random tests holds 50 tests with `$a$` and `$qs$` of size `$5000 \le size \le 20000$`, and `$-1E9 \le a[i] \le +1E9$`
```
### Example
```
a = [3, -8, 4, 5, -1], qs = [(1, 2), (1, 3), (2, 5), (4, 5), (3, 3)]
```
`subsequence_sums(a, qs)` should return:
```
[-13, -1, 11, 3, 4]
```
- For `$q=(1,2): 3 \cdot 1 + (-8) \cdot 2 = -13$`
- For `$q=(1,3): 3 \cdot 1 + (-8) \cdot 2 + 4 \cdot 3 = -1$`
- For `$q=(2,5): (-8) \cdot 1 + 4 \cdot 2 + 5 \cdot 3 + (-1) \cdot 4 = 11$`
- For `$q=(4,5): 5 \cdot 1 + (-1) \cdot 2 = 3$`
- For `$q=(3,3): 4 \cdot 1 = 4$`
$
| algorithms | # -*- coding:utf-8 -*-
# author : utoppia
# description : solutions for codewars.com
# updated at : 2022-02-03 16:19
# -----------------------------------------------------
# Kata UUID : 61e2edfeaf28c2001b57af98
# Title : Subsequence Sums II
# Kyu : 5
# Kata's Sensi: tonylicoding
# Tags : ['Algorithms', 'Mathematics', 'Numbers']
# Collection : 2
# Bookmarked : 8
# Solved : 28/31
# Language : Python
# ----------------------------------------------------
def subsequence_sums(a, qs):
"""
Let b[i] = a[i] + ... + a[n],
therefore for some query (l, r),
the answer is a[l] + a[l+1] * 2 + ... + a[r] * (r - l + 1),
which can be writen as (a[l] + ... + a[r]) + (a[l+1] + ... + a[r]) + ... + (a[r])
let c = a[r+1] + ... + a[n],
the equation upon can be expressed as
(b[l] - c) + (b[l+1] - c) + ... + (b[r] - c) = b[l] + ... + b[r] - c * (r - l + 1)
thus, we can make s[i] = a[1] + ... + a[i],
for some query (l, r), we can calculate c quickly as s[n] - s[r]
and make S[i] = b[1] + ... + b[i], therefor b[l] + ... + b[r] = S[r] - S[l-1],
ans = S[r] - S[l-1] - (s[n] - s[r]) * (r - l + 1)
problem has been solved!
"""
n = len(a)
b = [0 for _ in range(n + 1)]
tmp = 0
for i in range(n, 0, - 1):
tmp += a[i - 1]
b[i] = tmp
S = [0 for _ in range(n + 1)]
for i in range(1, n + 1):
S[i] = S[i - 1] + b[i]
s = [0 for _ in range(n + 1)]
for i in range(1, n + 1):
s[i] = s[i - 1] + a[i - 1]
ret = []
for l, r in qs:
ret . append(S[r] - S[l - 1] - (s[n] - s[r]) * (r - l + 1))
return ret
| Subsequence Sums II | 61e2edfeaf28c2001b57af98 | [
"Mathematics",
"Algorithms"
] | https://www.codewars.com/kata/61e2edfeaf28c2001b57af98 | 5 kyu |
## Setup
Start with an empty square grid that hypothetically goes on forever.
You will be given a list of points (`(x, y)` coordinates) where the initial stones are to be placed which all have a value of `1`.
For example input of `[(0, 0), (1, 0)]` is the initial setup corresponding to two stones side by side.
While an input of `[(0, 0), (2, -2)]` can be represented as if you imagine the x axis to increase to the right and y axis to increase to the top, and `(0, 0)` being the top left `1`:
```
1 0 0
0 0 0
0 0 1
```
## Rules
Your task is to place stones with increasing values starting from 2 (`2, 3, 4, ...`) with the rule, that a stone valued X can only be placed on an empty spot, where the sum of its eight neighbours equals to X.
```
1 0 0
0 X 0
0 0 1
```
In the above setup a stone with value `2` can be placed at position X.
```
1 0 0
0 0 Y
0 0 1
```
In the above setup a stone with value `2` can NOT be placed at position Y, because the sum of its neighbours is not `2` but only `1`.
After having placed the `2` you have to place `3` next and so on.
There is a numberphile video explaining these rules: https://www.youtube.com/watch?v=m4Uth-EaTZ8
## Task
Your task is given an initial setup of stones (all valued 1) to place stones with increasing values starting with 2 according to the above rules until you find the *highest* valued stone which cannot be placed on the board.
## Example
Given the above setup of `[(0, 0), (2, -2)]` it is possible to places stones valued 2, 3, 4 up to 16. But it is not possible to place the next stone valued 17.
```
9 5 10 11 0 0
0 4 1 0 0 0
12 8 3 2 0 16
0 0 0 6 1 15
0 0 13 7 14 0
```
So the solution is 17. | games | from collections import defaultdict
def neighbors(pos):
i, j = pos
return [(i + k, j + l) for k in range(- 1, 2) for l in range(- 1, 2) if k or l]
def first_impossible(elems) - > int:
memo, done = defaultdict(int), set(elems)
def update(p, n):
for pos in neighbors(p):
if pos not in done:
memo[pos] += n
def rec(n):
if not (coords := [k for k, v in memo . items() if v == n and k not in done]):
return n
res = 0
for p1 in coords:
done . add(p1)
update(p1, n)
res = max(res, rec(n + 1))
done . remove(p1)
update(p1, - n)
return res
for p in elems:
update(p, 1)
return rec(2)
| Stepping Stones Puzzle | 61e1b974eb372a001f719527 | [
"Games",
"Lists",
"Data Structures",
"Puzzles"
] | https://www.codewars.com/kata/61e1b974eb372a001f719527 | 4 kyu |
Welcome to the world of the National Football League!
In the NFL the Triple Crown is given when a receiver has the most receiving yards, the most receiving touchdowns and the most receptions in a single season.
This year Cooper Kupp managed to get it, however it is quite rare because the last one was in 2005 by Steve Smith.
Now you will receive a dictionary with the following keys (will always contain each):
- Cooper Kupp
- Justin Jefferson
- Davante Adams
Each key will have another dictionary as their values with the following keys:
- Receiving yards (value between 1500-2000)
- Receiving touchdowns (value between 10-20)
- Receptions (value between 90-120)
If one receiver has the most in each category you have to return his name. If there is no receiver with the most values in all categories you should return 'None of them'.
Example:
```python
{
'Cooper Kupp':
{
'Receiving yards': 1786,
'Receiving touchdowns': 18,
'Receptions': 117
},
'Justin Jefferson':
{
'Receiving yards': 1700,
'Receiving touchdowns': 17,
'Receptions': 115
},
'Davante Adams':
{
'Receiving yards': 1650,
'Receiving touchdowns': 16,
'Receptions': 110
}
}
# The output should be 'Cooper Kupp' since he has more receiving yards, more receiving touchdowns and more receptions as well
```
Example with two receivers sharing values in at least one category:
```python
{
'Cooper Kupp':
{
'Receiving yards': 1900,
'Receiving touchdowns': 18,
'Receptions': 117
},
'Justin Jefferson':
{
'Receiving yards': 1800,
'Receiving touchdowns': 17,
'Receptions': 116
},
'Davante Adams':
{
'Receiving yards': 1900,
'Receiving touchdowns': 18,
'Receptions': 110
}
}
# The output should be 'None of them' since they are tied on receiving yards and receiving touchdowns
```
| reference | def triple_crown(r):
for i in r:
if all(r[i][k] > r[j][k] for k in r[i] for j in r if i != j):
return i
return 'None of them'
| Triple Crown | 61e173ccbc916700267ef2ae | [
"Fundamentals"
] | https://www.codewars.com/kata/61e173ccbc916700267ef2ae | 7 kyu |
## Problem
The 'water pouring' problem (a.k.a 'two water jug' problem, or the even cooler 'Die Hard with a vengeance' puzzle) is a puzzle involving a finite number of water jugs with integer capacities (in this case, measured in litres). For this problem, we are using only two jugs. Initially, each jug is empty. We want to find a series of 'steps' to reach a state where ***one of the two jugs*** contain the desired quantity of water.
---
## Task
### Inputs:
1. Volume of the first jug, `a`
2. Volume of the second jug, `b` (`a<=b`)
3. The desired quantity of water, `n`
* Range of possible inputs `0<a,b,n<=100000`
### Outputs:
1. A list of tuples, with the state of the two jugs after every step starting from the **first step** . This need not be the 'path' with the minimum number of steps, but it will surely help for some random performance tests
### Valid 'Steps':
1. Emptying one jug and doing nothing with the other
2. Filling one jug and doing nothing with the other
3. Emptying one jug into the other
4. Using one jug to fill the other (refer to steps 2 and 6 in the example)
*In short, a valid step consists of pouring water from a source container (either of the two jugs or from the infinite source of water) to a destination (either of the two jugs or poured outside) until either the source container is empty or the destination container is full.*
---
### Example:
If `a=3, b=5, n=4`, the steps could be:
Start from `(0,0)`
1. Fill the second jug `(0,5)` *(Note that the first step is the first step with either of the jugs filled)*
2. Empty 3 litres from second jug to first jug `(3,2)`
3. Empty first jug `(0,2)`
4. Empty 2 litres from second jug to first jug `(2,0)`
5. Fill second jug `(2,5)`
6. Empty 1 litre from second jug to first jug `(3,4)`. We've reached a state where one of the jugs contains the desired quantity of water `n=4`.
*(The next step of emptying the first jug is redundant, but having the last step as (0,4) does not amount to a wrong answer)*
`(0,0)->(0,5)->(3,2)->(0,2)->(2,0)->(2,5)->(3,4)`
The function can return
```python
[(0,5), (3,2), (0,2), (2,0), (2,5), (3,4)]
```
---
### Note:
* Assume that you have an infinite supply of water to use.
* Both jugs start out empty, you can choose to fill either of them.
* If the goal cannot be reached in *one jug alone* return an empty list [].
* All inputs will be > 0
* [Learn more about this puzzle](https://en.wikipedia.org/wiki/Water_pouring_puzzle) | games | def wpp(a, b, n):
z = []
for i in range(b):
c = (a * i) % b
d = min(c + a, b)
z . append((a, c))
z . append((c + a - d, d))
if n in (c + a - d, d):
return z
if c + a >= b:
z . append((c + a - d, 0))
z . append((0, c + a - d))
return []
| Water pouring problem | 61de6142b31ff7000cc27e10 | [
"Algorithms",
"Performance",
"Puzzles",
"Graph Theory"
] | https://www.codewars.com/kata/61de6142b31ff7000cc27e10 | 5 kyu |
# Task
The function `fold_cube(number_list)` should return a boolean based on a net (given as a number list), if the net can fold into a cube.
Your code must be effecient and complete the tests within 500ms.
## Input
Imagine a net such as the one below.
```
@
@ @ @ @
@
```
Then, put it on the table with each `@` being one cell.
```
--------------------------
| 1 | 2 | 3 | 4 | 5 |
--------------------------
| 6 | 7 | 8 | 9 | 10 |
--------------------------
| 11 | 12 | 13 | 14 | 15 |
--------------------------
| 16 | 17 | 18 | 19 | 20 |
--------------------------
| 21 | 22 | 23 | 24 | 25 |
--------------------------
```
The number list for a net will be the numbers the net falls on when placed on the table. Note that the numbers in the list won't always be in order. The position of the net can be anywhere on the table, it can also be rotated.
For the example above, a possible number list will be `[1, 7, 8, 6, 9, 13]`.
## Output
`fold_cube` should return `True` if the net can be folded into a cube. `False` if not.
---
# Examples
### Example 1:
`number_list` = `[24, 20, 14, 19, 18, 9]`
Shape and rotation of net:
```
@
@
@ @ @
@
```
Displayed on the table:
```
--------------------------
| 1 | 2 | 3 | 4 | 5 |
--------------------------
| 6 | 7 | 8 | @ | 10 |
--------------------------
| 11 | 12 | 13 | @ | 15 |
--------------------------
| 16 | 17 | @ | @ | @ |
--------------------------
| 21 | 22 | 23 | @ | 25 |
--------------------------
```
So, `fold_cube([24, 20, 14, 19, 18, 9])` should return `True`.
### Example 2:
`number_list` = `[1, 7, 6, 17, 12, 16]`
Shape and rotation of net:
```
@
@ @
@
@ @
```
Displayed on the table:
```
--------------------------
| @ | 2 | 3 | 4 | 5 |
--------------------------
| @ | @ | 8 | 9 | 10 |
--------------------------
| 11 | @ | 13 | 14 | 15 |
--------------------------
| @ | @ | 18 | 19 | 20 |
--------------------------
| 21 | 22 | 23 | 24 | 25 |
--------------------------
```
`fold_cube([1, 7, 6, 17, 12, 16])` should return `False`.
#### *Translations appreciated!!*
---
### If you liked this kata.... check out these!
- [Folding a 4D Cube (Tesseract)](https://www.codewars.com/kata/5f3b561bc4a71f000f191ef7)
- [Wraping a net around a cube](https://www.codewars.com/kata/5f4af9c169f1cd0001ae764d)
| games | def fold_cube(nums):
return expand(nums . pop(), set(nums), 1, 2, 3) == {1, 2, 3, - 1, - 2, - 3}
def expand(val, nums, x, y, z):
dirs = {z}
for num in nums . copy():
if abs(val - num) not in (1, 5) or {val % 5, num % 5} == {0, 1}:
continue
nums . discard(num)
diff = val - num
sign = diff / / abs(diff)
nx, ny, nz = (x, z * sign, - y *
sign) if abs(diff) == 1 else (- z * sign, y, x * sign)
dirs |= expand(num, nums, nx, ny, nz)
return dirs
| Folding a cube | 5f2ef12bfb6eb40019cc003e | [
"Puzzles",
"Algorithms"
] | https://www.codewars.com/kata/5f2ef12bfb6eb40019cc003e | 5 kyu |
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