# Minimum sum of differences with an element in an array

• Difficulty Level : Easy
• Last Updated : 18 May, 2021

Given an array, we need to find the sum of elements of an array after changing the element as arr[i] will become abs(arr[i]-x) where x is an array element.

Examples:

```Input  : {2, 5, 1, 7, 4}
Output : 9
We get minimum sum when we choose
x = 4. The minimum sum is
abs(2-4) + abs(5-4) + abs(1-4) + (7-4)
abs(4-4) = 9

Input  : {5, 11, 14, 10, 17, 15}
Output : 20
We can either choose x = 14 or x = 11```

The idea is based on fact that middle element would cause minimum sum of differences. When there are even number of elements, we can take any of the middle two elements. We can verify this fact by taking few examples.

Below is the implementation of above idea:

## C++

 `// C++ program to find minimum sum of absolute``// differences with an array element.``#include``using` `namespace` `std;``    ``// function to find min sum after operation``    ``int` `absSumDidd(``int` `a[],``int` `n)``    ``{``        ``// Sort the array``        ``sort(a,a+n);``        ` `        ``// Pick middle value``        ``int` `midValue = a[(``int``)(n / 2)];` `        ``// Sum of absolute differences.``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < n; i++) {``            ``sum = sum + ``abs``(a[i] - midValue);``        ``}``        ` `        ``return` `sum;    ``    ``}` `    ``// Driver Code``    ``int` `main()``    ``{``        ``int` `arr[] = { 5, 11, 14, 10, 17, 15 };``        ``int` `n=``sizeof``(arr)/``sizeof``(arr[0]);``        ``cout<< absSumDidd(arr,n);    ``    ``}``    ``// Contributed by mits`

## Java

 `// Java program to find minimum sum of absolute``// differences with an array element.``import` `java.lang.*;``import` `java.util.*;` `public` `class` `GFG {` `    ``// function to find min sum after operation``    ``static` `int` `absSumDidd(``int` `a[])``    ``{``        ``// Sort the array``        ``Arrays.sort(a);``        ` `        ``// Pick middle value``        ``int` `midValue = a[a.length / ``2``];` `        ``// Sum of absolute differences.``        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < a.length; i++) {``            ``sum = sum + Math.abs(a[i] - midValue);``        ``}``       ` `        ``return` `sum;      ``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``5``, ``11``, ``14``, ``10``, ``17``, ``15` `};``        ``System.out.print(absSumDidd(arr));       ``    ``}``    ``// Contributed by Saurav Jain``}`

## Python3

 `# Python3 program to find minimum sum of``# absolute differences with an array element.` `# function to find min sum after operation``def` `absSumDidd(a, n):``    ` `    ``# Sort the array``    ``a.sort()``    ` `    ``# Pick middle value``    ``midValue ``=` `a[(``int``)(n ``/``/` `2``)]` `    ``# Sum of absolute differences.``    ``sum` `=` `0``    ``for` `i ``in` `range``(n):``        ``sum` `=` `sum` `+` `abs``(a[i] ``-` `midValue)``    ``return` `sum` `# Driver Code``arr ``=` `[``5``, ``11``, ``14``, ``10``, ``17``, ``15``]``n ``=` `len``(arr)``print``(absSumDidd(arr, n))` `# This code is contributed``# by sahishelangia`

## C#

 `// C# program to find minimum sum of absolute``// differences with an array element.``using` `System;`  ` ``class` `GFG {``  ` `    ``// function to find min sum after operation``    ``static` `int` `absSumDidd(``int` `[]a)``    ``{``        ``// Sort the array``        ``Array.Sort(a);``          ` `        ``// Pick middle value``        ``int` `midValue = a[a.Length / 2];``  ` `        ``// Sum of absolute differences.``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < a.Length; i++) {``            ``sum = sum + Math.Abs(a[i] - midValue);``        ``}``         ` `        ``return` `sum;       ``    ``}``  ` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 5, 11, 14, 10, 17, 15 };``        ``Console.Write(absSumDidd(arr));        ``    ``}``    ``// Contributed by Subhadeep``}`

## PHP

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## Javascript

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Output:

`20`

Time Complexity: O(n Log n)
We can further optimize above solution to O(n) by using linear time algorithm for median finding.

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