Minimize the sum of the squares of the sum of elements of each group the array is divided into

Given an array consisting of even number of elements, the task is to divide the array into M group of elements (every group must contain at least 2 elements) such that the sum of the squares of the sums of each group is minimized i.e., 
(sum_of_elements_of_group1)² + (sum_of_elements_of_group2)² + (sum_of_elements_of_group3)² + (sum_of_elements_of_group4)² + ..... + (sum_of_elements_of_groupM)²
Examples: 
 

Input: arr[] = {5, 8, 13, 45, 6, 3} 
Output: 2824 
Groups can be (3, 45), (5, 13) and (6, 8) 
(3 + 45)² + (5 + 13)² + (6 + 8)² = 48² + 18² + 14² = 2304 + 324 + 196 = 2824
Input: arr[] = {53, 28, 143, 5} 
Output: 28465 
 

Approach: Our final sum depends on two factors: 
 

1. Sum of the elements of each group.
2. The sum of squares of all such groups.

If we minimize both the factors mentioned above, we can minimize the result. To minimize the second factor we should make groups of minimum size i.e. just two elements. To minimize first factor we can pair smallest number with largest number, second smallest number to second largest number and so on.
Below is the implementation of the above approach: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the minimized sum
unsigned long long findAnswer(int n, 
                       vector<int>& arr)
{

    // Sort the array to pair the elements
    sort(arr.begin(), arr.end());

    // Variable to hold the answer
    unsigned long long sum = 0;

    // Pair smallest with largest, second
    // smallest with second largest, and 
    // so on
    for (int i = 0; i < n / 2; ++i) {
        sum += (arr[i] + arr[n - i - 1])
               * (arr[i] + arr[n - i - 1]);
    }

    return sum;
}

// Driver code
int main()
{
    std::vector<int> arr = { 53, 28, 143, 5 };
    int n = arr.size();
    cout << findAnswer(n, arr);
}

// Java implementation of the approach
import java.util.*;

class GFG 
{

    // Function to return the minimized sum
    static int findAnswer(int n, int[] arr)
    {

        // Sort the array to pair the elements
        Arrays.sort(arr);

        // Variable to hold the answer
        int sum = 0;

        // Pair smallest with largest, second
        // smallest with second largest, and 
        // so on
        for (int i = 0; i < n / 2; ++i) 
        {
            sum += (arr[i] + arr[n - i - 1])
                    * (arr[i] + arr[n - i - 1]);
        }

        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = {53, 28, 143, 5};
        int n = arr.length;
        System.out.println(findAnswer(n, arr));
    }
}

// This code has been contributed by 29AjayKumar

# Python 3 implementation of the approach

# Function to return the minimized sum
def findAnswer(n, arr):
    
    # Sort the array to pair the elements
    arr.sort(reverse = False)

    # Variable to hold the answer
    sum = 0

    # Pair smallest with largest, second
    # smallest with second largest, and 
    # so on
    for i in range(int(n / 2)):
        sum += ((arr[i] + arr[n - i - 1]) * 
                (arr[i] + arr[n - i - 1]))

    return sum

# Driver code
if __name__ == '__main__':
    arr = [53, 28, 143, 5]
    n = len(arr)
    print(findAnswer(n, arr))

# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;

class GFG
{
    
// Function to return the minimized sum
static int findAnswer(int n, int []arr)
{

    // Sort the array to pair the elements
    Array.Sort(arr);

    // Variable to hold the answer
    int sum = 0;

    // Pair smallest with largest, second
    // smallest with second largest, and 
    // so on
    for (int i = 0; i < n / 2; ++i) 
    {
        sum += (arr[i] + arr[n - i - 1])
            * (arr[i] + arr[n - i - 1]);
    }

    return sum;
}

// Driver code
static void Main()
{
    int []arr = { 53, 28, 143, 5 };
    int n = arr.Length;
    Console.WriteLine(findAnswer(n, arr));
}
}

// This code is contributed by mits

<?php
// PHP implementation of the approach

// Function to return the minimized sum
function findAnswer($n, $arr)
{

    // Sort the array to pair the elements
    sort($arr);

    // Variable to hold the answer
    $sum = 0;

    // Pair smallest with largest, second
    // smallest with second largest, and 
    // so on
    for ($i = 0; $i < $n / 2; ++$i) 
    {
        $sum += ($arr[$i] + $arr[$n - $i - 1]) * 
                ($arr[$i] + $arr[$n - $i - 1]);
    }

    return $sum;
}

// Driver code
$arr = array( 53, 28, 143, 5);
$n = count($arr);
echo findAnswer($n, $arr);

// This code is contributed by chandan_jnu
?>

// Javascript implementation of the approach

// Function to return the minimized sum
function findAnswer(n, arr)
{

    // Sort the array to pair the elements
    arr.sort((a, b) => a - b);

    // Variable to hold the answer
    let sum = 0;

    // Pair smallest with largest, second
    // smallest with second largest, and
    // so on
    for (let i = 0; i < Math.floor(n / 2); ++i)
    {
        sum += (arr[i] + arr[n - i - 1]) *
                (arr[i] + arr[n - i - 1]);
    }

    return sum;
}

// Driver code
let arr = new Array( 53, 28, 143, 5);
let n = arr.length;
document.write(findAnswer(n, arr));


// This code is contributed by _saurabh_jaiswal

28465

Time Complexity: O(nlogn), used for sorting the array
Auxiliary Space: O(1), as no extra space is used