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: 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 26Input: arr[] = {7, 2, 13, 4, 25, 8}
Output: 2845
Approach: The task is to maximize the sum of a2 + b2 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 arrayint 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 codeint 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 approachimport 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 approachusing 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))
Auxiliary Space: O(1)
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