Given an integer array **arr[]**, the task is to divide this array into two non-empty subsets such that the sum of the square of the sum of both the subsets is maximum and sizes of both the subsets must not differ by more than 1.**Examples:**

Input:arr[] = {1, 2, 3}Output:26Explanation:

Sum of Subset Pairs are as follows

(1)^{2}+ (2 + 3)^{2}= 26

(2)^{2}+ (1 + 3)^{2}= 20

(3)^{2}+ (1 + 2)^{2}= 18

Maximum among these is 26, Therefore the required sum is 26Input:arr[] = {7, 2, 13, 4, 25, 8}Output:2845

**Approach:** The task is to maximize the sum of **a ^{2} + b^{2}** where

**a**and

**b**are the sum of the two subsets and

**a + b = C**(constant), i.e., the sum of the entire array. The maximum sum can be achieved by sorting the array and dividing the first

**N/2 – 1**smaller elements in one subset and the rest

**N/2 + 1**elements in the other subset. In this way, the sum can be maximized while keeping the difference in size at most 1.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the maximum sum of the` `// square of the sum of two subsets of an array` `int` `maxSquareSubsetSum(` `int` `* A, ` `int` `N)` `{` ` ` `// Initialize variables to store` ` ` `// the sum of subsets` ` ` `int` `sub1 = 0, sub2 = 0;` ` ` `// Sorting the array` ` ` `sort(A, A + N);` ` ` `// Loop through the array` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// Sum of the first subset` ` ` `if` `(i < (N / 2) - 1)` ` ` `sub1 += A[i];` ` ` `// Sum of the second subset` ` ` `else` ` ` `sub2 += A[i];` ` ` `}` ` ` `// Return the maximum sum of` ` ` `// the square of the sum of subsets` ` ` `return` `sub1 * sub1 + sub2 * sub2;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 7, 2, 13, 4, 25, 8 };` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << maxSquareSubsetSum(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` `class` `GFG` `{` ` ` ` ` `// Function to return the maximum sum of the` ` ` `// square of the sum of two subsets of an array` ` ` `static` `int` `maxSquareSubsetSum(` `int` `[]A, ` `int` `N)` ` ` `{` ` ` `// Initialize variables to store` ` ` `// the sum of subsets` ` ` `int` `sub1 = ` `0` `, sub2 = ` `0` `;` ` ` ` ` `// Sorting the array` ` ` `Arrays.sort(A);` ` ` ` ` `// Loop through the array` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++)` ` ` `{` ` ` ` ` `// Sum of the first subset` ` ` `if` `(i < (N / ` `2` `) - ` `1` `)` ` ` `sub1 += A[i];` ` ` ` ` `// Sum of the second subset` ` ` `else` ` ` `sub2 += A[i];` ` ` `}` ` ` ` ` `// Return the maximum sum of` ` ` `// the square of the sum of subsets` ` ` `return` `sub1 * sub1 + sub2 * sub2;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `7` `, ` `2` `, ` `13` `, ` `4` `, ` `25` `, ` `8` `};` ` ` `int` `N = arr.length;` ` ` ` ` `System.out.println(maxSquareSubsetSum(arr, N));` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Python3

`# Python3 implementation of the approach` `# Function to return the maximum sum of the` `# square of the sum of two subsets of an array` `def` `maxSquareSubsetSum(A, N) :` ` ` `# Initialize variables to store` ` ` `# the sum of subsets` ` ` `sub1 ` `=` `0` `; sub2 ` `=` `0` `;` ` ` ` ` `# Sorting the array` ` ` `A.sort();` ` ` `# Loop through the array` ` ` `for` `i ` `in` `range` `(N) :` ` ` `# Sum of the first subset` ` ` `if` `(i < (N ` `/` `/` `2` `) ` `-` `1` `) :` ` ` `sub1 ` `+` `=` `A[i];` ` ` `# Sum of the second subset` ` ` `else` `:` ` ` `sub2 ` `+` `=` `A[i];` ` ` `# Return the maximum sum of` ` ` `# the square of the sum of subsets` ` ` `return` `sub1 ` `*` `sub1 ` `+` `sub2 ` `*` `sub2;` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `7` `, ` `2` `, ` `13` `, ` `4` `, ` `25` `, ` `8` `];` ` ` `N ` `=` `len` `(arr);` ` ` `print` `(maxSquareSubsetSum(arr, N));` `# This code is contributed by AnkitRai01` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` ` ` `// Function to return the maximum sum of the` ` ` `// square of the sum of two subsets of an array` ` ` `static` `int` `maxSquareSubsetSum(` `int` `[]A, ` `int` `N)` ` ` `{` ` ` `// Initialize variables to store` ` ` `// the sum of subsets` ` ` `int` `sub1 = 0, sub2 = 0;` ` ` ` ` `// Sorting the array` ` ` `Array.Sort(A);` ` ` ` ` `// Loop through the array` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `{` ` ` ` ` `// Sum of the first subset` ` ` `if` `(i < (N / 2) - 1)` ` ` `sub1 += A[i];` ` ` ` ` `// Sum of the second subset` ` ` `else` ` ` `sub2 += A[i];` ` ` `}` ` ` ` ` `// Return the maximum sum of` ` ` `// the square of the sum of subsets` ` ` `return` `sub1 * sub1 + sub2 * sub2;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[]arr = { 7, 2, 13, 4, 25, 8 };` ` ` `int` `N = arr.Length;` ` ` ` ` `Console.WriteLine(maxSquareSubsetSum(arr, N));` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` `// javascript implementation of the approach` `// Creating the bblSort function` ` ` `function` `bblSort(arr){` ` ` ` ` `for` `(` `var` `i = 0; i < arr.length; i++){` ` ` ` ` `// Last i elements are already in place ` ` ` `for` `(` `var` `j = 0; j < ( arr.length - i -1 ); j++){` ` ` ` ` `// Checking if the item at present iteration` ` ` `// is greater than the next iteration` ` ` `if` `(arr[j] > arr[j+1]){` ` ` ` ` `// If the condition is true then swap them` ` ` `var` `temp = arr[j]` ` ` `arr[j] = arr[j + 1]` ` ` `arr[j+1] = temp` ` ` `}` ` ` `}` ` ` `}` ` ` `// return the sorted array` ` ` `return` `arr;` `}` ` ` `// Function to return the maximum sum of the` ` ` `// square of the sum of two subsets of an array` ` ` `function` `maxSquareSubsetSum(A , N) {` ` ` `// Initialize variables to store` ` ` `// the sum of subsets` ` ` `var` `sub1 = 0, sub2 = 0;` ` ` `// Sorting the array` ` ` `A = bblSort(A);` ` ` `// Loop through the array` ` ` `for` `(i = 0; i < N; i++) {` ` ` `// Sum of the first subset` ` ` `if` `(i < (N / 2) - 1)` ` ` `sub1 += A[i];` ` ` `// Sum of the second subset` ` ` `else` ` ` `sub2 += A[i];` ` ` `}` ` ` `// Return the maximum sum of` ` ` `// the square of the sum of subsets` ` ` `return` `sub1 * sub1 + sub2 * sub2;` ` ` `}` ` ` `// Driver code` ` ` ` ` `var` `arr = [ 7, 2, 13, 4, 25, 8 ];` ` ` `var` `N = arr.length;` ` ` `document.write(maxSquareSubsetSum(arr, N));` `// This code is contributed by todaysgaurav` `</script>` |

**Output:**

2845

**Time Complexity:** O(N*log(N))

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