Deal the cards to the players

Question

Tonight is card game night! You are the dealer and your task is to write a program to deal the cards to the players.

Given an array of cards and the number of players, you need to split the array of cards into a hand for each player.

Rules

Your program will receive an non-empty array A , as well as a non-zero positive integer n. The array should then be split into n hands. If the length of the string isn't divisible by n any leftover cards at the end should be distributed as evenly as possible.

• If n==1, you will need to return an array of array with A as it's only element

• If n is greater than the length of A, you will need to return every hand and an empty hand. if n = 4 and array A = [1,2,3], you should return [[1],[2],[3]] or [[1],[2],[3],[]]. You are free to handle the empty hand with empty, undefined or null.

• The array can contain any type rather than a number.

• You should not change the order of the array while dealing. For example if n = 2 and A= [1,2,3], any result rather than [[1,3],[2]] will be invalid.

Test Cases

n   A               Output

1   [1,2,3,4,5,6]   [[1,2,3,4,5,6]]
2   [1,2,3,4,5,6]   [[1,3,5],[2,4,6]]
3   [1,2,3,4,5,6]   [[1,4],[2,5],[3,6]]
4   [1,2,3,4,5,6]   [[1,5],[2,6],[3],[4]]
7   [1,2,3,4,5,6]   [[1],[2],[3],[4],[5],[6]] // or [[1],[2],[3],[4],[5],[6],[]]

Demo Program

def deal(cards, n):
	i = 0
	players = [[] for _ in range(n)]
	for card in cards:
		players[i % n].append(card)
		i += 1
	return players

hands = deal([1,2,3,4,5,6], 2)

print(hands)

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This is code-golf, so you the shortest bytes of each language will be the winner.

Inspired from Create chunks from array by chau giang

Answer 1

05AB1E, 3 1 byte

Saved 2 bytes thanks to Adnan

ι

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Explanation

ι  # uninterleave

Does exactly what the challenge asks for

Answer 2

R, 46 25 bytes

function(A,n)split(A,1:n)

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splits A into groups defined by 1:n, recycling 1:n until it matches length with A.

Answer 3

Perl 6, 33 24 bytes

->\b{*.classify:{$++%b}}

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Anonymous curried code block that takes a number and returns a Whatever lambda that takes a list and returns a list of lists. This takes the second option when given a number larger than the length of lists, e.g. f(4)([1,2,3]) returns [[1],[2],[3]]

Explanation:

->\b{                  }  # Anonymous code block that takes a number
     *                    # And returns a Whatever lambda
      .classify           # That groups by
               :{$++%b}   # The index modulo the number

Answer 4

Wolfram Language (Mathematica), 28 bytes

(s=#;GatherBy[#2,#~Mod~s&])&

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Answer 5

Japt, 2 bytes

Takes the array as the first input.

óV

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Answer 6

Jelly, 6 2 bytes

sZ

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Thanks to @JonathanAllan for saving 4 bytes

Answer 7

Python 2, 37 bytes

Code:

lambda x,n:[x[i::n]for i in range(n)]

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Answer 8

Haskell, 39 bytes

import Data.Lists
(transpose.).chunksOf

Note: Data.Lists is from the third-party library lists, which is not on Stackage and hence will not appear on Hoogle.

Answer 9

J, 13, 11, 10, 9 bytes

(|#\)</.]

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how (previous explanation, fundamentally the same)

] </.~ (| #\)
  </.~          NB. box results of grouping
]               NB. the right arg by...
         |      NB. the remainders of dividing...
       [        NB. the left arg into...
           #\   NB. the length of each prefix of...
              ] NB. the right arg,
                NB. aka, the integers 1 thru
                NB. the length of the right arg

Answer 10

Charcoal, 9 bytes

ＩＥθ✂ηιＬηθ

Try it online! Link is to verbose version of code. Takes input in the order [n, A] and outputs each value on its own line and each hand double-spaced from the previous. Explanation:

  θ         First input `n`
 Ｅ          Map over implicit range
    η       Second input `A`
   ✂        Sliced
     ι      Starting at current index
      Ｌη    Ending at length of `A`
        θ   Taking every `n`th element
Ｉ           Cast to string
            Implicitly print

Answer 11

Kotlin, 53 51 49 bytes

{a,n->(0..n-1).map{a.slice(it..a.size-1 step n)}}

The old, incorrect solution only worked for divisors of the array length. I'm certain this can be golfed down.

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Answer 12

Wolfram Language (Mathematica), 25 bytes

#2[[i;;;;#]]~Table~{i,#}&

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Answer 13

TSQL, 44 bytes

-- @       : table containing the input 
-- column c: value of the card, 
-- column a: position on the card in the deck
-- @n      : number of players

DECLARE @ table(a int identity(0,1), c varchar(9))
DECLARE @n int = 4

INSERT @ values('1a'),('2c'),('3e'),('4g'),('5i'),('6k')

SELECT string_agg(c,',')FROM @ GROUP BY a%@n

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Answer 14

JavaScript (Node.js), 51 bytes

A=>n=>A.map((x,i)=>e[i%=n]=[...e[i]||[],x],e=[])&&e

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JavaScript (Node.js), 53 bytes

A=>n=>g=(i=n)=>i?[...g(--i),A.filter(_=>i--%n==0)]:[]

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Answer 15

Ruby, 81 bytes

def s a,n;a.each_with_index.inject(([[]]*n).map(&:dup)){|b,(c,d)|b[d%n]<<c;b};end

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Answer 16

APL+WIN 26 or 31 bytes

If individual hands can be represented as columns of a 2D matrix then 26 bytes if an array of arrays then add 5 bytes.

(l,n)⍴((l←⌈(⍴a)÷n)×n←⎕)↑a←⎕

Try it online! ourtesy of Dyalog Classic

or

⊂[1](l,n)⍴((l←⌈(⍴a)÷n)×n←⎕)↑a←⎕

Try it online! Courtesy of Dyalog Classic

Explanation:

a←⎕ prompt for array of cards

((l←⌈(⍴a)÷n)×n←⎕)↑ prompt for integer, pad a with zeros to given even hands

(l,n)⍴ create 2D matrix with each column representing each hand

⊂[1] if required convert to nested vector - APL array of arrays

Answer 17

MathGolf, 9 bytes

\ô_í\%q╞;

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Explanation

\           swap top elements (pops both input onto stack)
 ô          start block of length 6
  _         duplicate TOS (will duplicate the list)
   í        get total number of iterations of for loop (the other input)
    \       swap top elements
     %      modulo (picks every n:th item of the list
      q     print without newline
       ╞    discard from left of string/array (makes the next player pick cards starting with the next in the deck)
        ;   discard TOS (removes some junk in the end)

Answer 18

Java (JDK), 90 bytes

A->n->{var o="";for(int h=0,i;h<n;o+="\n")for(i=h++;i<A.length;i+=n)o+=" "+A[i];return o;}

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Thanks Olivier Grégoire for the lambda and better incrementing while iterating.

Answer 19

PHP, 85 83 82 bytes

function($a,$n){while($x<$n)$c[]=array_column(array_chunk($a,$n),+$x++);return$c;}

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This will not be the shortest entry, but I thought it'd be fun to try and do it using PHP array function built-ins. Result: long.

Output

1   [1,2,3,4,5,6]   [[1,2,3,4,5,6]]
2   [1,2,3,4,5,6]   [[1,3,5],[2,4,6]]
3   [1,2,3,4,5,6]   [[1,4],[2,5],[3,6]]
4   [1,2,3,4,5,6]   [[1,5],[2,6],[3],[4]]
7   [1,2,3,4,5,6]   [[1],[2],[3],[4],[5],[6],[]]
5   ["9C","QD","2S","4H","6D","AS","9D","TH","5C"]  [["9C","AS"],["QD","9D"],["2S","TH"],["4H","5C"],["6D"]]

Answer 20

Perl 5 -la, 66 56 bytes

$,=<>;map{push@$_,shift@F}1..$,while@F;say"@$_"for 1..$,

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Answer 21

C# (Visual C# Interactive Compiler), 43 bytes

a=>b=>{int i=0;return a.GroupBy(_=>i++%b);}

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Answer 22

C (gcc), 5 bytes

The compiler flag -Df= (req. leading space) fulfills specification. f(n_cards,n_hands,card_ptr) evaluates to a pointer to a list of hands.

Explanation

In C, it is common practice to implement lists of lists as a single interleaved array, when the number of lists remains constant but all lists can be extended. For example, in this case of dealing cards, it is more common for more cards to get added to each hand than more hands to be added, so it would be reasonable to implement a list of hands as an interleaved list. Coincidentally, the "deck" is such a list, and thus we return the parameter unmodified.

This challenge probably should've been sandboxed.