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

• Difficulty Level : Easy
• Last Updated : 13 Jun, 2022

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++

 `// C++ implementation of the approach``#include ``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

 `// 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`

## Python3

 `# 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#

 `// 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

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## Javascript

 `// 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`

Output:

`28465`

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

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