Choose k array elements such that difference of maximum and minimum is minimized


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.


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:

  1. Sort the Array.
  2. Calculate the maximum(k numbers) – minimum(k numbers) for each group of k consecutive integers.
  3. 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.
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;



Time Complexity: O(nlogn).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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