Given an array of n integers where each value represents number of chocolates in a packet. Each packet can have 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 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 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 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: 10 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 maximum and minimum elements in it. Finally return minimum difference.

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

## 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. int first = 0, last = 0; for (int i=0; i+m-1<n; i++) { int diff = arr[i+m-1] - arr[i]; if (diff < min_diff) { min_diff = diff; first = i; last = i + m - 1; } } return (arr[last] - arr[first]); } 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. int first = 0, last = 0; for (int i = 0; i + m - 1 < n; i++) { int diff = arr[i+m-1] - arr[i]; if (diff < min_diff) { min_diff = diff; first = i; last = i + m - 1; } } return (arr[last] - arr[first]); } /* 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.

Output:

Minimum difference is 10

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

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