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.

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) - min(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 remains the same. If we perform the same procedure for other p like numbers, 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 the implementation of above idea : 

C++

// 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;
}

Java

// Java program to find minimum difference
// of maximum and minimum of K number.
import java.util.*;

class GFG {

// Return minimum difference of
// maximum and minimum of k
// elements of arr[0..n-1].
static int minDiff(int arr[], int n, int k) {
    int result = Integer.MAX_VALUE;

    // Sorting the array.
    Arrays.sort(arr);

    // Find minimum value among
    // all K size subarray.
    for (int i = 0; i <= n - k; i++)
    result = Math.min(result, arr[i + k - 1] - arr[i]);

    return result;
}

// Driver code
public static void main(String[] args) {
    int arr[] = {10, 100, 300, 200, 1000, 20, 30};
    int n = arr.length;
    int k = 3;

    System.out.println(minDiff(arr, n, k));
}
}

// This code is contributed by Anant Agarwal.

Python3

# Python program to find minimum
# difference of maximum
# and minimum of K number.

# Return minimum difference
# of maximum and minimum
# of k elements of arr[0..n-1].
def minDiff(arr,n,k):
    result = +2147483647

    # Sorting the array.
    arr.sort()

    # Find minimum value among
    # all K size subarray.
    for i in range(n-k+1):
        result = int(min(result, arr[i+k-1] - arr[i]))

    return result

# Driver code

arr= [10, 100, 300, 200, 1000, 20, 30]
n =len(arr)
k = 3

print(minDiff(arr, n, k))

# This code is contributed
# by Anant Agarwal.

C#

// C# program to find minimum
// difference of maximum and
// minimum of K number.
using System;

class GFG {

// Return minimum difference of
// maximum and minimum of k
// elements of arr[0..n - 1].
static int minDiff(int []arr, int n,
                   int k)
{
    int result = int.MaxValue;

    // Sorting the array.
    Array.Sort(arr);

    // Find minimum value among
    // all K size subarray.
    for (int i = 0; i <= n - k; i++)
    result = Math.Min(result, arr[i + k - 1] - arr[i]);

    return result;
}

// Driver code
public static void Main() {
    int []arr = {10, 100, 300, 200, 1000, 20, 30};
    int n = arr.Length;
    int k = 3;

    Console.WriteLine(minDiff(arr, n, k));
}
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find minimum
// difference of maximum and
// minimum of K number.

// Return minimum difference
// of maximum and minimum
// of k elements of arr[0..n-1].
function minDiff($arr, $n, $k)
{
    $INT_MAX = 2147483647;
    $result = $INT_MAX ;

    // Sorting the array.
    sort($arr , $n);
    sort($arr);

    // Find minimum value among
    // all K size subarray.
    for ($i = 0; $i <= $n - $k; $i++)
        $result = min($result, $arr[$i + $k - 1] -
                                       $arr[$i]);
    return $result;
}

    // Driver Code
    $arr = array(10, 100, 300, 200, 1000, 20, 30);
    $n = sizeof($arr);
    $k = 3;
    echo minDiff($arr, $n, $k);

// This code is contributed by nitin mittal.
?>

Javascript

<script>
// javascript program to find minimum difference
// of maximum and minimum of K number.

    // Return minimum difference of
    // maximum and minimum of k
    // elements of arr[0..n-1].
    function minDiff(arr , n , k) {
        var result = Number.MAX_VALUE;

        // Sorting the array.
        arr.sort((a,b)=>a-b);

        // Find minimum value among
        // all K size subarray.
        for (i = 0; i <= n - k; i++)
            result = Math.min(result, arr[i + k - 1] - arr[i]);

        return result;
    }

    // Driver code

        var arr = [ 10, 100, 300, 200, 1000, 20, 30 ];
        var n = arr.length;
        var k = 3;

        document.write(minDiff(arr, n, k));

// This code contributed by gauravrajput1
</script>

Output

20

Time Complexity: O(nlogn).
Auxiliary Space: O(1)

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