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[1], arr[3]) 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 <bits/stdc++.h>
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

<script>

// JavaScript program for the above approach

// Function to find the minimum value
// of the sum of absolute difference
// between all pairs of arrays
function minSumDifference(ar, n) {
  // Stores the sum of array elements
  let sum = 0;

  // Find the sum of array element
  for (let i = 0; i < n; i++) sum += ar[i];

  // Store the value of sum%N
  let rem = sum % n;

  // Return the resultant value
  return rem * (n - rem);
}

// Driver Code

let arr = [3, 6, 8, 5, 2, 1, 11, 7, 10, 4];
let N = arr.length;
document.write(minSumDifference(arr, N));

</script>

Output:

21

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