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 = 6Input : 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:
Java
// Java program for Chocolate Distribution // problemimport 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 code 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}; // Number of students int m = 7; 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.
Auxiliary Space: O(1)
As constant extra space is used.
Please refer complete article on Chocolate Distribution Problem for more details!
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