# Minimize sum of absolute difference between all pairs of array elements by decrementing and incrementing pairs by 1

Given an array arr[] ( 1-based indexing ) consisting of N integers, the task is to find the minimum sum of the absolute difference between all pairs of array elements by decrementing and incrementing any pair of elements by 1 any number of times.

Examples:

Input: arr[] = {1, 2, 3}
Output: 0
Explanation:
Modify the array elements by  performing the following operations:

• Choose the pairs of element (arr, arr) and incrementing and decrementing the pairs modifies the array to {2, 2, 2}.

After the above operations, the sum of the absolute differences is  |2 – 2| + |2 – 2| + |2 – 2| = 0. Therefore, print 0.

Input: arr[] = {0, 1, 0, 1}
Output: 4

Approach: The given problem can be solved by using a Greedy Approach. It can be observed that to minimize the sum of the absolute difference between every pair of array elements arr[], the idea to make every array element closed to each other. Follow the steps below to solve the problem:

• Find the sum of the array elements arr[] and store it in a variable, say sum.
• Now, if the value of sum % N is 0, then print 0 as all the array elements can be made equal and the resultant value of the expression is always 0. Otherwise, find the value of sum % N and store it in a variable, say R.
• Now, if all the array elements are sum/N, then we can make R number of array elements as 1 and the rest of the array elements as 0 to minimize the resultant value.
• After the above steps, the minimum sum of the absolute difference is given by R*(N – R).

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the minimum value` `// of the sum of absolute difference` `// between all pairs of arrays` `int` `minSumDifference(``int` `ar[], ``int` `n)` `{` `    ``// Stores the sum of array elements` `    ``int` `sum = 0;`   `    ``// Find the sum of array element` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``sum += ar[i];`   `    ``// Store the value of sum%N` `    ``int` `rem = sum % n;`   `    ``// Return the resultant value` `    ``return` `rem * (n - rem);` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 3, 6, 8, 5, 2,` `                  ``1, 11, 7, 10, 4 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``cout << minSumDifference(arr, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `class` `GFG {`   `    ``// Function to find the minimum value` `    ``// of the sum of absolute difference` `    ``// between all pairs of arrays` `    ``public` `static` `int` `minSumDifference(``int` `ar[], ``int` `n) {` `        ``// Stores the sum of array elements` `        ``int` `sum = ``0``;`   `        ``// Find the sum of array element` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``sum += ar[i];`   `        ``// Store the value of sum%N` `        ``int` `rem = sum % n;`   `        ``// Return the resultant value` `        ``return` `rem * (n - rem);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String args[]) {` `        ``int``[] arr = { ``3``, ``6``, ``8``, ``5``, ``2``, ``1``, ``11``, ``7``, ``10``, ``4` `};` `        ``int` `N = arr.length;` `        ``System.out.println(minSumDifference(arr, N));`   `    ``}` `}`   `// This code is contributed by gfgking.`

## Python3

 `# Python 3 program for the above approach`   `# Function to find the minimum value` `# of the sum of absolute difference` `# between all pairs of arrays` `def` `minSumDifference(ar, n):` `    ``# Stores the sum of array elements` `    ``sum` `=` `0`   `    ``# Find the sum of array element` `    ``for` `i ``in` `range``(n):` `        ``sum` `+``=` `ar[i]`   `    ``# Store the value of sum%N` `    ``rem ``=` `sum` `%` `n`   `    ``# Return the resultant value` `    ``return` `rem ``*` `(n ``-` `rem)`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``3``, ``6``, ``8``, ``5``, ``2``, ``1``, ``11``, ``7``, ``10``, ``4``]` `    ``N ``=` `len``(arr)` `    ``print``(minSumDifference(arr, N))` `    `  `    ``# This code is contributed by ipg2016107.`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{`   `// Function to find the minimum value` `// of the sum of absolute difference` `// between all pairs of arrays` `public` `static` `int` `minSumDifference(``int``[] ar, ``int` `n) ` `{` `    `  `    ``// Stores the sum of array elements` `    ``int` `sum = 0;`   `    ``// Find the sum of array element` `    ``for``(``int` `i = 0; i < n; i++)` `        ``sum += ar[i];`   `    ``// Store the value of sum%N` `    ``int` `rem = sum % n;`   `    ``// Return the resultant value` `    ``return` `rem * (n - rem);` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``int``[] arr = { 3, 6, 8, 5, 2, ` `                  ``1, 11, 7, 10, 4 };` `    ``int` `N = arr.Length;` `    `  `    ``Console.Write(minSumDifference(arr, N));` `}` `}`   `// This code is contributed by sanjoy_62`

## Javascript

 ``

Output:

`21`

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

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