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 = 3Output:Minimum Difference is 2Explanation:

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 = 5Output:Minimum Difference is 6Explanation:

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 = 7Output:Minimum Difference is 10Explanation:

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` `# 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` `?>` |

**Output**

Minimum difference is 10

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

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