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# Minimize the sum of squares of sum of N/2 paired formed by N numbers

Given N numbers(N is an even number). Divide the N numbers into N/2 pairs in such a way that the sum of squares of the sum of numbers in pairs is minimal. The task is to print the minimal sum.

Examples:

```Input: a[] = {8, 5, 2, 3}
Output: 164
Divide them into two groups of {2, 8} and {3, 5}
to give (2+8)2 + (3+5)2 = 164

Input: a[] = {1, 1, 1, 2, 2, 2}
Output: 27 ```

Approach: The task is to minimize the sum of squares, hence we divide the smallest and largest number in the first group and the second smallest and second largest in the second group, and so on till N/2 groups are formed. Add the numbers and store the sum of squares of them which will be the final answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to minimize the sum``// of squares of sum of numbers``// of N/2 groups of N numbers` `#include ``using` `namespace` `std;` `// Function that returns the minimize sum``// of square of sum of numbers of every group``int` `findMinimal(``int` `a[], ``int` `n)``{``    ``// Sort the array elements``    ``sort(a, a + n);` `    ``int` `sum = 0;` `    ``// Iterate for the first half of array``    ``for` `(``int` `i = 0; i < n / 2; i++)``        ``sum += (a[i] + a[n - i - 1])``                ``* (a[i] + a[n - i - 1]);` `    ``return` `sum;``}` `// Driver Code``int` `main()``{``    ``int` `a[] = { 8, 5, 2, 3 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);`` ` `    ``cout << findMinimal(a, n);`` ` `    ``return` `0;``}`

## Java

 `// Java program to minimize the sum``// of squares of sum of numbers``// of N/2 groups of N numbers``import` `java.util.Arrays;` `class` `GFG``{` `    ``// Function that returns the minimize sum``    ``// of square of sum of numbers of every group``    ``static` `int` `findMinimal(``int` `[]a, ``int` `n)``    ``{``        ``// Sort the array elements``        ``Arrays.sort(a);``    ` `        ``int` `sum = ``0``;``    ` `        ``// Iterate for the first half of array``        ``for` `(``int` `i = ``0``; i < n / ``2``; i++)``            ``sum += (a[i] + a[n - i - ``1``]) *``                ``(a[i] + a[n - i - ``1``]);``    ` `        ``return` `sum;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String str[])``    ``{``        ``int` `[]a = { ``8``, ``5``, ``2``, ``3` `};``        ``int` `n = a.length;``        ``System.out.println(findMinimal(a, n));``    ``}``}` `// This code is contributed by Ryuga`

## Python 3

 `# Python 3 program to minimize the sum``# of squares of sum of numbers``# of N/2 groups of N numbers` `# Function that returns the minimize sum``# of square of sum of numbers of every group``def` `findMinimal(a, n):` `    ``# Sort the array elements``    ``a.sort()` `    ``sum` `=` `0` `    ``# Iterate for the first half of array``    ``for` `i ``in` `range``( n ``/``/` `2``):``        ``sum` `+``=` `((a[i] ``+` `a[n ``-` `i ``-` `1``]) ``*``                ``(a[i] ``+` `a[n ``-` `i ``-` `1``]))` `    ``return` `sum` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``a ``=` `[ ``8``, ``5``, ``2``, ``3` `]``    ``n ``=` `len``(a)` `    ``print``(findMinimal(a, n))` `# This code is contributed by ita_c`

## C#

 `// C# program to minimize the sum``// of squares of sum of numbers``// of N/2 groups of N numbers``using` `System;` `class` `GFG``{` `// Function that returns the minimize sum``// of square of sum of numbers of every group``static` `int` `findMinimal(``int` `[]a, ``int` `n)``{``    ``// Sort the array elements``    ``Array.Sort(a);` `    ``int` `sum = 0;` `    ``// Iterate for the first half of array``    ``for` `(``int` `i = 0; i < n / 2; i++)``        ``sum += (a[i] + a[n - i - 1]) *``               ``(a[i] + a[n - i - 1]);` `    ``return` `sum;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]a = { 8, 5, 2, 3 };``    ``int` `n = a.Length;` `    ``Console.Write(findMinimal(a, n));``}``}` `// This code is contributed``// by Akanksha Rai`

## PHP

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## Javascript

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Output

`164`

Complexity Analysis:

• Time Complexity: O(N*logN), as we are using an inbuilt sort function that will cost N*logN time. Where N is the number of elements in the array.
• Auxiliary Space: O(1), as we are not using any extra space.

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