C++ Program For 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: 

1. Each student gets one packet.
2. 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;
}

// Driver code
int main()
{
    int arr[] = {12, 4,  7,  9,  2,  23, 25,
                 41, 30, 40, 28, 42, 30, 44,
                 48, 43, 50 };

    // Number of students
    int m = 7;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Minimum difference is " <<
             findMinDiff(arr, n, m);
    return 0;
}

Output:Minimum difference is 10

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

Auxiliary Space: O(1)

As constant extra space is used.

Please refer complete article on Chocolate Distribution Problem for more details!