Given an array of **n** integers and a positive number **k**. We are allowed to take any k integers from the given array. The task is to find the minimum possible value of the difference between maximum and minimum of K numbers.

Examples:

Input : arr[] = {10, 100, 300, 200, 1000, 20, 30} k = 3 Output : 20 20 is the minimum possible difference between any maximum and minimum of any k numbers. Given k = 3, we get the result 20 by selecting integers {10, 20, 30}. max(10, 20, 30) - max(10, 20, 30) = 30 - 10 = 20. Input : arr[] = {1, 2, 3, 4, 10, 20, 30, 40, 100, 200}. k = 4 Output : 3

The idea is to sort the array and choose k continuous integers. Why continuous? let the chosen k integers be arr[0], arr[1],,..arr[r], arr[r+x]…, arr[k-1], all in increasing order but not continuous in the sorted array. This means there exists an integer p which lies between arr[r] and arr[r+x],. So if p is included and arr[0] is removed, then the new difference will be arr[r] – arr[1] whereas old difference was arr[r] – arr[0]. And we know arr[0] <= arr[1] <= ..<= arr[k-1] so minimum difference reduces or remain same. If we perform same procedure for other p like number, we get the minimum difference. Algorithm to solve the problem:

- Sort the Array.
- Calculate the maximum(k numbers) – minimum(k numbers) for each group of k consecutive integers.
- Return minimum of all values obtained in step 2.

Below is C++ implementation of this approach:

// C++ program to find minimum difference of maximum // and minimum of K number. #include<bits/stdc++.h> using namespace std; // Return minimum difference of maximum and minimum // of k elements of arr[0..n-1]. int minDiff(int arr[], int n, int k) { int result = INT_MAX; // Sorting the array. sort(arr, arr + n); // Find minimum value among all K size subarray. for (int i=0; i<=n-k; i++) result = min(result, arr[i+k-1] - arr[i]); return result; } // Driven Program int main() { int arr[] = {10, 100, 300, 200, 1000, 20, 30}; int n = sizeof(arr)/sizeof(arr[0]); int k = 3; cout << minDiff(arr, n, k) << endl; return 0; }

Output:

20

**Time Complexity: **O(nlogn).

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