Daily Challenge #82 - English Beggars

Given an array of values and a number of beggars, you are supposed to return an array with the sum of what each beggar brings home, assuming they all take regular turns, from the first to the last.

For example: [1,2,3,4,5] for 2 beggars will return a result of [9,6], as the first one takes [1,3,5], the second collects [2,4].

The same array with 3 beggars would have in turn have produced a better outcome for the second beggar: [5,7,3], as they will respectively take [1,4], [2,5] and [3].

Also note that not all beggars have to take the same amount of "offers", meaning that the length of the array is not necessarily a multiple of n; length can be even shorter, in which case the last beggars will, of course, take nothing (0).

---

This challenge comes from GiacomoSorbi on CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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Comments

[Tōm]

function reduceAmount(arr, amount, index) {
    arr[index % arr.length] += amount;
    return arr;
}

function beggars(amounts, beggarsNumber) {
    const sums = [...Array(beggarsNumber)].map(_ => 0);
    return amounts.reduce(reduceAmount, sums);
}

beggars([1,2,3,4,5], 2) // [9, 6]

[Amin]

I'm not sure as I can't test it out here but I think that this line

const sums = [...Array(beggarsNumber)].map(_ => 0);

Could have been written as

const sums = [...Array(beggarsNumber)].fill(0);

Even if it is a small change, I find it easier to read.

[Tōm]

Thank you for your response, it can also be changed as :

const sums = Array(10).fill(0);

I forgot this function, and it is really helpful and permits to avoid the hackish code I wrote above.

Thanks!

[K.V.Harish]

My solution in js

const beggarsCollection = (collection, numOfBeggars) => {
  const sum = Array(numOfBeggars).fill(0);
  collection.forEach((value, index) => {
    sum[index % numOfBeggars] += value;
  });
  return sum;
};

[LorenzoTa]

A simple perl solution consuming the array

use strict;
use warnings;

my @values = (1..5);
my @beggars;
$#beggars = $ARGV[0]//1;
my $ind = 0;


$beggars[$ind]+= shift @values and $ind >= $#beggars ? $ind = 0 : $ind++ while @values;

Or, using Tye::Cycle from CPAN a very clean solution:

use Tie::Cycle;

my @values = (1..5);
my @beggars;
$#beggars = $ARGV[0]//1;

tie my $ind, 'Tie::Cycle', [ 0..$#beggars ];

$beggars[$ind]+= $_ for @values;

[E. Choroba]

Going functional in Perl:

#!/usr/bin/perl
use warnings;
use strict;

use List::Util      qw{ sum };
use List::MoreUtils qw{ part };

sub beggars {
    my ($money, $number_of_beggars) = @_;
    my $i = 0;
    [ map sum(@$_), part { $i++ % $number_of_beggars } @$money ];
}

use Test::More tests => 2;
is_deeply beggars([1, 2, 3, 4, 5], 2), [9,6];
is_deeply beggars([1, 2, 3, 4, 5], 3), [5, 7, 3];

[DeLiXx]

An computation-on-demand approach in js

function beggars(arr, count) {
    return (i) => {
        let loot = 0;
        for(; i < arr.length; i += count) loot += arr[i];
        return loot;
    }
}
let a = beggars([1,2,3,4,5],2);

[Amin]

Elm

module EnglishBeggars exposing (englishBeggars)


currentBeggar distribution turn donation =
    donation -- (0, 1)
        |> Tuple.first -- 0
        |> modBy distribution -- modBy 0 = 0, modBy 1 = 1, modBy 2 = 0
        |> (==) turn -- 0 == 0 = True, 1 == 0 = False, 0 == 0 = True


distribute donations distribution turn =
    donations -- [1, 2, 3, 4, 5]
        |> List.indexedMap Tuple.pair -- [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
        |> List.filter (currentBeggar distribution turn) -- [(0, 1), (2, 3), (4, 5)]
        |> List.map Tuple.second -- [1, 3, 5]
        |> List.sum -- 9


englishBeggars : List Int -> Int -> List Int
englishBeggars donations distribution =
    List.range 0 (distribution - 1) -- [0, 1]
        |> List.map (distribute donations distribution) -- [9, 6]

Tests

module EnglishBeggarsTest exposing (suite)

import EnglishBeggars exposing (englishBeggars)
import Expect exposing (equal)
import Test exposing (Test, describe, test)


suite : Test
suite =
    describe "English beggars"
        [ test "Should return [9, 6] for a donation of [1, 2, 3, 4, 5] between 2 beggars" <|
            \_ -> englishBeggars [ 1, 2, 3, 4, 5 ] 2 |> equal [ 9, 6 ]
        , test "Should return [5, 7, 3] for a donation of [1, 2, 3, 4, 5] between 3 beggars" <|
            \_ -> englishBeggars [ 1, 2, 3, 4, 5 ] 3 |> equal [ 5, 7, 3 ]
        , test "Should return [1, 2, 3, 0] for a donation of [1, 2, 3] between 4 beggars" <|
            \_ -> englishBeggars [ 1, 2, 3 ] 4 |> equal [ 1, 2, 3, 0 ]
        ]

[Peter]

In Go!

amts below could also be a []float64 if required. All posted examples were with ints so that is what I used.

func beggars(amts []int, beggarCount int) []int {
    // defaults to slice of 0's
    sums := make([]int, beggarCount)
    for idx, amt := range amts {
        beggarId := idx % beggarCount
        sums[beggarId] += amt
    }
    return sums
}

Go Playground Example

[Amin]

Could you remind me what is the concept of point free style? I know I heard it somewhere when playing with Haskell but I can't remember haha!