# Chocolate Distribution Problem

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 :

1. Each student gets one packet.
2. 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.
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

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.

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