Chocolate Distribution Problem

Given an array of N integers where each value represents the number of chocolates in a packet. Each packet can have a variable number of chocolates. There are m students, the task is to distribute chocolate packets such that: 

• Each student gets one packet.
• The difference between the number of chocolates in the packet with maximum chocolates and the packet with minimum chocolates given to the students is minimum.

Examples:

Input : arr[] = {7, 3, 2, 4, 9, 12, 56} , m = 3 
Output: Minimum Difference is 2 
Explanation:
We have seven packets of chocolates and we need to pick three packets for 3 students 
If we pick 2, 3 and 4, we get the minimum difference between maximum and minimum packet sizes.

Input : arr[] = {3, 4, 1, 9, 56, 7, 9, 12} , m = 5 
Output: Minimum Difference is 6 

Input : arr[] = {12, 4, 7, 9, 2, 23, 25, 41, 30, 40, 28, 42, 30, 44, 48, 43, 50} , m = 7 
Output: Minimum Difference is 10 

Naive Approach for Chocolate Distribution Problem

The idea is to generate all subsets of size m of arr[0..n-1]. For every subset, find the difference between the maximum and minimum elements in it. Finally, return the minimum difference.

Efficient Approach for Chocolate Distribution Problem

The idea is based on the observation that to minimize the difference, we must choose consecutive elements from a sorted packet. We first sort the array arr[0..n-1], then find the subarray of size m with the minimum difference between the last and first elements.

Illustration:

Below is the illustration of example arr[] = [ 7,3,2,4,9,12,56 ] , m = 3

Follow the steps mentioned below to implement the approach:

• Initially sort the given array. And declare a variable to store the minimum difference, and initialize it to INT_MAX. Let the variable be min_diff.
• Find the subarray of size m such that the difference between the last (maximum in case of sorted) and first (minimum in case of sorted) elements of the subarray is minimum.
• We will run a loop from 0 to (n-m), where n is the size of the given array and m is the size of the subarray.
• We will calculate the maximum difference with the subarray and store it in diff = arr[highest index] – arr[lowest index]
• Whenever we get a diff less than the min_diff, we will update the min_diff with diff.

Below is the implementation of the above approach:

C++

// C++ program to solve chocolate distribution
// problem
#include <bits/stdc++.h>
using namespace std;

// arr[0..n-1] represents sizes of packets
// m is number of students.
// Returns minimum difference between maximum
// and minimum values of distribution.
int findMinDiff(int arr[], int n, int m)
{
    // if there are no chocolates or number
    // of students is 0
    if (m == 0 || n == 0)
        return 0;

    // Sort the given packets
    sort(arr, arr + n);

    // Number of students cannot be more than
    // number of packets
    if (n < m)
        return -1;

    // Largest number of chocolates
    int min_diff = INT_MAX;

    // Find the subarray of size m such that
    // difference between last (maximum in case
    // of sorted) and first (minimum in case of
    // sorted) elements of subarray is minimum.

    for (int i = 0; i + m - 1 < n; i++) {
        int diff = arr[i + m - 1] - arr[i];
        if (diff < min_diff)
            min_diff = diff;
    }
    return min_diff;
}

int main()
{
    int arr[] = { 12, 4,  7,  9,  2,  23, 25, 41, 30,
                  40, 28, 42, 30, 44, 48, 43, 50 };
    int m = 7; // Number of students
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Minimum difference is "
         << findMinDiff(arr, n, m);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program to solve chocolate distribution problem
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>

// Compare function for qsort
int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}

// arr[0..n-1] represents sizes of packets
// m is number of students.
// Returns minimum difference between maximum
// and minimum values of distribution.
int findMinDiff(int arr[], int n, int m)
{
    // if there are no chocolates or number
    // of students is 0
    if (m == 0 || n == 0)
        return 0;

    // Sort the given packets
    qsort(arr, n, sizeof(int), cmpfunc);

    // Number of students cannot be more than
    // number of packets
    if (n < m)
        return -1;

    // Largest number of chocolates
    int min_diff = INT_MAX;

    // Find the subarray of size m such that
    // difference between last (maximum in case
    // of sorted) and first (minimum in case of
    // sorted) elements of subarray is minimum.

    for (int i = 0; i + m - 1 < n; i++) {
        int diff = arr[i + m - 1] - arr[i];
        if (diff < min_diff)
            min_diff = diff;
    }
    return min_diff;
}

int main()
{
    int arr[] = { 12, 4,  7,  9,  2,  23, 25, 41, 30,
                  40, 28, 42, 30, 44, 48, 43, 50 };
    int m = 7; // Number of students
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Minimum difference is %d",
           findMinDiff(arr, n, m));
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// JAVA Code For Chocolate Distribution
// Problem
import java.util.*;

class GFG {

    // arr[0..n-1] represents sizes of
    // packets. m is number of students.
    // Returns minimum difference between
    // maximum and minimum values of
    // distribution.
    static int findMinDiff(int arr[], int n, int m)
    {
        // if there are no chocolates or
        // number of students is 0
        if (m == 0 || n == 0)
            return 0;

        // Sort the given packets
        Arrays.sort(arr);

        // Number of students cannot be
        // more than number of packets
        if (n < m)
            return -1;

        // Largest number of chocolates
        int min_diff = Integer.MAX_VALUE;

        // Find the subarray of size m
        // such that difference between
        // last (maximum in case of
        // sorted) and first (minimum in
        // case of sorted) elements of
        // subarray is minimum.

        for (int i = 0; i + m - 1 < n; i++) {
            int diff = arr[i + m - 1] - arr[i];
            if (diff < min_diff)
                min_diff = diff;
        }
        return min_diff;
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int arr[] = { 12, 4,  7,  9,  2,  23, 25, 41, 30,
                      40, 28, 42, 30, 44, 48, 43, 50 };

        int m = 7; // Number of students

        int n = arr.length;
        System.out.println("Minimum difference is "
                           + findMinDiff(arr, n, m));
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to solve
# chocolate distribution
# problem


# arr[0..n-1] represents sizes of packets
# m is number of students.
# Returns minimum difference between maximum
# and minimum values of distribution.
def findMinDiff(arr, n, m):

    # if there are no chocolates or number
    # of students is 0
    if (m == 0 or n == 0):
        return 0

    # Sort the given packets
    arr.sort()

    # Number of students cannot be more than
    # number of packets
    if (n < m):
        return -1

    # Largest number of chocolates
    min_diff = arr[n-1] - arr[0]

    # Find the subarray of size m such that
    # difference between last (maximum in case
    # of sorted) and first (minimum in case of
    # sorted) elements of subarray is minimum.
    for i in range(len(arr) - m + 1):
        min_diff = min(min_diff,  arr[i + m - 1] - arr[i])

    return min_diff


# Driver Code
if __name__ == "__main__":
    arr = [12, 4, 7, 9, 2, 23, 25, 41,
           30, 40, 28, 42, 30, 44, 48,
           43, 50]
    m = 7  # Number of students
    n = len(arr)
    print("Minimum difference is", findMinDiff(arr, n, m))

# This code is contributed by Smitha

C#

// C# Code For Chocolate Distribution
// Problem
using System;

class GFG {

    // arr[0..n-1] represents sizes of
    // packets. m is number of students.
    // Returns minimum difference between
    // maximum and minimum values of
    // distribution.
    static int findMinDiff(int[] arr, int n, int m)
    {

        // if there are no chocolates or
        // number of students is 0
        if (m == 0 || n == 0)
            return 0;

        // Sort the given packets
        Array.Sort(arr);

        // Number of students cannot be
        // more than number of packets
        if (n < m)
            return -1;

        // Largest number of chocolates
        int min_diff = int.MaxValue;

        // Find the subarray of size m
        // such that difference between
        // last (maximum in case of
        // sorted) and first (minimum in
        // case of sorted) elements of
        // subarray is minimum.

        for (int i = 0; i + m - 1 < n; i++) {
            int diff = arr[i + m - 1] - arr[i];

            if (diff < min_diff)
                min_diff = diff;
        }

        return min_diff;
    }

    /* Driver program to test above function */
    public static void Main()
    {
        int[] arr = { 12, 4,  7,  9,  2,  23, 25, 41, 30,
                      40, 28, 42, 30, 44, 48, 43, 50 };

        int m = 7; // Number of students

        int n = arr.Length;

        Console.WriteLine("Minimum difference is "
                          + findMinDiff(arr, n, m));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to solve
// chocolate distribution
// problem

// arr[0..n-1] represents
// sizes of packets m is
// number of students.
// Returns minimum difference
// between maximum and minimum
// values of distribution.
function findMinDiff($arr, $n, $m)
{
    // if there are no
    // chocolates or number
    // of students is 0
    if ($m == 0 || $n == 0)
        return 0;

    // Sort the given packets
    sort($arr);

    // Number of students
    // cannot be more than
    // number of packets
    if ($n < $m)
    return -1;

    // Largest number
    // of chocolates
    $min_diff = PHP_INT_MAX;

    // Find the subarray of size
    // m such that difference
    // between last (maximum in
    // case of sorted) and first
    // (minimum in case of sorted)
    // elements of subarray is minimum.

    for ($i = 0;
         $i + $m - 1 < $n; $i++)
    {
        $diff = $arr[$i + $m - 1] -
                $arr[$i];
        if ($diff < $min_diff)
            $min_diff = $diff;
    }
    return $min_diff;
}

// Driver Code
$arr = array(12, 4, 7, 9, 2, 23,
             25, 41, 30, 40, 28,
             42, 30, 44, 48, 43, 50);

$m = 7; // Number of students
$n = sizeof($arr);
echo "Minimum difference is ",
    findMinDiff($arr, $n, $m);

// This code is contributed by ajit
?>

Javascript

<script>

// Javascript Code For Chocolate
// Distribution Problem

// arr[0..n-1] represents sizes of
// packets. m is number of students.
// Returns minimum difference between
// maximum and minimum values of
// distribution.
function findMinDiff(arr, n, m)
{

    // If there are no chocolates or
    // number of students is 0
    if (m == 0 || n == 0)
        return 0;

    // Sort the given packets
    arr.sort(function(a, b){return a - b});

    // Number of students cannot be
    // more than number of packets
    if (n < m)
        return -1;

    // Largest number of chocolates
    let min_diff = Number.MAX_VALUE;

    // Find the subarray of size m
    // such that difference between
    // last (maximum in case of
    // sorted) and first (minimum in
    // case of sorted) elements of
    // subarray is minimum.
    for(let i = 0; i + m - 1 < n; i++)
    {
        let diff = arr[i + m - 1] - arr[i];

        if (diff < min_diff)
            min_diff = diff;
    }
    return min_diff;
}

// Driver code
let arr = [ 12, 4, 7, 9, 2, 23, 25,
            41, 30, 40, 28, 42, 30,
            44, 48, 43, 50 ];

// Number of students
let m = 7;
let n = arr.length;

document.write("Minimum difference is " +
               findMinDiff(arr, n, m));

// This code is contributed by divyesh072019

</script>

Output

Minimum difference is 10

Time Complexity: O(N*log(N))
Auxiliary Space: O(1)