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 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
Explanation:
The set goes like 3,4,7,9,9 and the output
is 9-3 = 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
Explanation:
We need to pick 7 packets. We pick 40, 41,
42, 44, 48, 43 and 50 to minimize difference
between maximum and minimum.

Source: Flipkart Interview Experience

A simple solution 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.
An efficient solution 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.

Below image is a dry run of the above approach:

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;
}

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
 
import sys;
 
# 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 = sys.maxsize
 
    # 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.
    i=0
    while(i+m-1<n ):
     
        diff = arr[i+m-1] - arr[i]
        if (diff < min_diff):
            min_diff = diff
         
        i+=1
         
    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
?>

Output

Minimum difference is 10

Time Complexity: O(n Log n) as we apply sorting before subarray search.

This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

C++

Java

Python3

C#

PHP

Recommended Posts: