# Maximize difference between maximum and minimum array elements after K operations

• Last Updated : 08 Apr, 2021

Given an array arr[] of size N and a positive integer K, the task is to find the maximum difference between the largest element and the smallest element in the array by incrementing or decrementing array elements by 1, K times.

Examples:

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Input: arr[] = {7, 7, 7, 7}, K = 1
Output: 14
Explanation: Decrementing the value of arr and incrementing the value of arr by 7 modifies arr[] = {0, 7, 7, 14}. Therefore, the maximum difference between the largest element and the smallest element of the array is 14

Input: arr[] = {0, 0, 0, 0, 0}, K = 2
Output: 0
Explanation: Since all array elements are 0, decrementing any array element makes that element less than 0. Therefore, the required output is 0.

Approach: Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to find the maximum difference``// between the maximum and minimum in the``// array after K operations``int` `maxDiffLargSmallOper(``int` `arr[],``                         ``int` `N, ``int` `K)``{``    ``// Stores maximum difference between``    ``// largest  and smallest array element``    ``int` `maxDiff = 0;` `    ``// Sort the array in descending order``    ``sort(arr, arr + N, greater<``int``>());` `    ``// Traverse the array arr[]``    ``for` `(``int` `i = 0; i <= min(K, N - 1);``         ``i++) {` `        ``// Update maxDiff``        ``maxDiff += arr[i];``    ``}` `    ``return` `maxDiff;``}` `// Driver Code``int` `main()``{` `    ``int` `arr[] = { 7, 7, 7, 7 };``    ``int` `N = ``sizeof``(arr)``            ``/ ``sizeof``(arr);``    ``int` `K = 1;``    ``cout << maxDiffLargSmallOper(arr, N, K);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{``    ` `// Reverse array``static` `int``[] reverse(``int` `a[])``{``    ``int` `i, n = a.length, t;``    ``for``(i = ``0``; i < n / ``2``; i++)``    ``{``        ``t = a[i];``        ``a[i] = a[n - i - ``1``];``        ``a[n - i - ``1``] = t;``    ``}``    ``return` `a;``}` `// Function to find the maximum difference``// between the maximum and minimum in the``// array after K operations``static` `int` `maxDiffLargSmallOper(``int` `arr[],``                                ``int` `N, ``int` `K)``{``    ` `    ``// Stores maximum difference between``    ``// largest  and smallest array element``    ``int` `maxDiff = ``0``;` `    ``// Sort the array in descending order``    ``Arrays.sort(arr);``    ``arr = reverse(arr);``    ` `    ``// Traverse the array arr[]``    ``for``(``int` `i = ``0``; i <= Math.min(K, N - ``1``); i++)``    ``{``        ` `        ``// Update maxDiff``        ``maxDiff += arr[i];``    ``}``    ` `    ``return` `maxDiff;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``7``, ``7``, ``7``, ``7` `};``    ``int` `N = arr.length;``    ``int` `K = ``1``;``    ` `    ``System.out.print(maxDiffLargSmallOper(arr, N, K));``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find the maximum difference``# between the maximum and minimum in the``# array after K operations``def` `maxDiffLargSmallOper(arr, N, K):``    ` `    ` `    ``# Stores maximum difference between``    ``# largest  and smallest array element``    ``maxDiff ``=` `0``;``    ` `    ``# Sort the array in descending order``    ``arr.sort(reverse ``=` `True``);``    ` `    ` `    ``# Traverse the array arr[]``    ``for` `i  ``in`  `range``(``min``(K ``+` `1``, N)):``        ` `        ``# Update maxDiff``        ``maxDiff ``+``=` `arr[i];``    ` `    ``return` `maxDiff;` `# Driver Code``if` `__name__ ``=``=` `"__main__"``: ` `    ``arr ``=` `[ ``7``, ``7``, ``7``, ``7` `];``    ``N ``=` `len``(arr)``    ``K ``=` `1``;``    ``print``(maxDiffLargSmallOper(arr, N, K));`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{` `// Function to find the maximum difference``// between the maximum and minimum in the``// array after K operations``static` `int` `maxDiffLargSmallOper(``int` `[]arr, ``int` `N,``                                ``int` `K)``{``    ` `    ``// Stores maximum difference between``    ``// largest and smallest array element``    ``int` `maxDiff = 0;` `    ``// Sort the array in descending order``    ``Array.Sort(arr);``    ``Array.Reverse(arr);``    ` `    ``// Traverse the array arr[]``    ``for``(``int` `i = 0; i <= Math.Min(K, N - 1); i++)``    ``{``        ` `        ``// Update maxDiff``        ``maxDiff += arr[i];``    ``}``    ``return` `maxDiff;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[] arr = ``new` `int``[]{ 7, 7, 7, 7 };``    ``int` `N = arr.Length;``    ``int` `K = 1;``    ` `    ``Console.Write(maxDiffLargSmallOper(arr, N, K));``}``}` `// This code is contributed by mohit kumar 29`

## Javascript

 ``
Output:
`14`

Time Complexity: O(N * log(N))
Auxiliary Space: O(1)

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