将数组分成两个子集，使得两个子集之和的平方和最大

给定一个整数数组arr[] ，任务是将该数组划分为两个非空子集，使得两个子集之和的平方和最大，并且两个子集的大小相差不得超过 1 .
例子：

Input: arr[] = {1, 2, 3} 
Output: 26 
Explanation: 
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 26
Input: arr[] = {7, 2, 13, 4, 25, 8} 
Output: 2845 
 

方法：任务是最大化a ² + b ²的总和，其中a和b是两个子集的总和， a + b = C (常数)，即整个数组的总和。最大和可以通过对数组进行排序并将前N/2 – 1 个较小元素划分为一个子集中的元素，并将其余N/2 + 1 个元素划分为另一个子集中的元素来实现。通过这种方式，可以在保持大小差异最多为 1 的情况下最大化和。
下面是上述方法的实现：

C++

// C++ implementation of the approach
 
#include 
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

// 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

输出：

2845

时间复杂度： O(N*log(N))

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