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 <bits/stdc++.h> 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[0]); 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
<?php // PHP 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 function findMinimal( $a , $n ) { // Sort the array elements sort( $a ); $sum = 0; // Iterate for the first half of array for ( $i = 0; $i < $n / 2; $i ++) $sum += ( $a [ $i ] + $a [ $n - $i - 1]) * ( $a [ $i ] + $a [ $n - $i - 1]); return $sum ; } // Driver Code $a = array (8, 5, 2, 3 ); $n = sizeof( $a ); echo findMinimal( $a , $n ); // This code is contributed by ajit ?> |
Javascript
<script> // java script 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 function findMinimal(a, n) { // Sort the array elements a.sort(); let sum = 0; // Iterate for the first half of array for (let i = 0; i < n / 2; i++) sum += (a[i] + a[n - i - 1]) * (a[i] + a[n - i - 1]); return sum; } // Driver Code let a = [8,5,2,3]; let n = a.length; document.write( findMinimal(a, n)); // This code is contributed by sravan kumar G </script> |
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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