Minimize the sum of the squares of the sum of elements of each group the array is divided into
Given an array consisting of even number of elements, the task is to divide the array into M group of elements (every group must contain at least 2 elements) such that the sum of the squares of the sums of each group is minimized i.e.,
(sum_of_elements_of_group1)2 + (sum_of_elements_of_group2)2 + (sum_of_elements_of_group3)2 + (sum_of_elements_of_group4)2 + ….. + (sum_of_elements_of_groupM)2
Examples:
Input: arr[] = {5, 8, 13, 45, 6, 3}
Output: 2824
Groups can be (3, 45), (5, 13) and (6, 8)
(3 + 45)2 + (5 + 13)2 + (6 + 8)2 = 482 + 182 + 142 = 2304 + 324 + 196 = 2824
Input: arr[] = {53, 28, 143, 5}
Output: 28465
Approach: Our final sum depends on two factors:
- Sum of the elements of each group.
- The sum of squares of all such groups.
If we minimize both the factors mentioned above, we can minimize the result. To minimize the second factor we should make groups of minimum size i.e. just two elements. To minimize first factor we can pair smallest number with largest number, second smallest number to second largest number and so on.
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 minimized sum unsigned long long findAnswer( int n, vector< int >& arr) { // Sort the array to pair the elements sort(arr.begin(), arr.end()); // Variable to hold the answer unsigned long long sum = 0; // Pair smallest with largest, second // smallest with second largest, and // so on for ( int i = 0; i < n / 2; ++i) { sum += (arr[i] + arr[n - i - 1]) * (arr[i] + arr[n - i - 1]); } return sum; } // Driver code int main() { std::vector< int > arr = { 53, 28, 143, 5 }; int n = arr.size(); cout << findAnswer(n, arr); } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the minimized sum static int findAnswer( int n, int [] arr) { // Sort the array to pair the elements Arrays.sort(arr); // Variable to hold the answer int sum = 0 ; // Pair smallest with largest, second // smallest with second largest, and // so on for ( int i = 0 ; i < n / 2 ; ++i) { sum += (arr[i] + arr[n - i - 1 ]) * (arr[i] + arr[n - i - 1 ]); } return sum; } // Driver code public static void main(String[] args) { int [] arr = { 53 , 28 , 143 , 5 }; int n = arr.length; System.out.println(findAnswer(n, arr)); } } // This code has been contributed by 29AjayKumar |
Python3
# Python 3 implementation of the approach # Function to return the minimized sum def findAnswer(n, arr): # Sort the array to pair the elements arr.sort(reverse = False ) # Variable to hold the answer sum = 0 # Pair smallest with largest, second # smallest with second largest, and # so on for i in range ( int (n / 2 )): sum + = ((arr[i] + arr[n - i - 1 ]) * (arr[i] + arr[n - i - 1 ])) return sum # Driver code if __name__ = = '__main__' : arr = [ 53 , 28 , 143 , 5 ] n = len (arr) print (findAnswer(n, arr)) # This code is contributed by # Surendra_Gangwar |
C#
// C# implementation of the approach using System; class GFG { // Function to return the minimized sum static int findAnswer( int n, int []arr) { // Sort the array to pair the elements Array.Sort(arr); // Variable to hold the answer int sum = 0; // Pair smallest with largest, second // smallest with second largest, and // so on for ( int i = 0; i < n / 2; ++i) { sum += (arr[i] + arr[n - i - 1]) * (arr[i] + arr[n - i - 1]); } return sum; } // Driver code static void Main() { int []arr = { 53, 28, 143, 5 }; int n = arr.Length; Console.WriteLine(findAnswer(n, arr)); } } // This code is contributed by mits |
PHP
<?php // PHP implementation of the approach // Function to return the minimized sum function findAnswer( $n , $arr ) { // Sort the array to pair the elements sort( $arr ); // Variable to hold the answer $sum = 0; // Pair smallest with largest, second // smallest with second largest, and // so on for ( $i = 0; $i < $n / 2; ++ $i ) { $sum += ( $arr [ $i ] + $arr [ $n - $i - 1]) * ( $arr [ $i ] + $arr [ $n - $i - 1]); } return $sum ; } // Driver code $arr = array ( 53, 28, 143, 5); $n = count ( $arr ); echo findAnswer( $n , $arr ); // This code is contributed by chandan_jnu ?> |
Javascript
// Javascript implementation of the approach // Function to return the minimized sum function findAnswer(n, arr) { // Sort the array to pair the elements arr.sort((a, b) => a - b); // Variable to hold the answer let sum = 0; // Pair smallest with largest, second // smallest with second largest, and // so on for (let i = 0; i < Math.floor(n / 2); ++i) { sum += (arr[i] + arr[n - i - 1]) * (arr[i] + arr[n - i - 1]); } return sum; } // Driver code let arr = new Array( 53, 28, 143, 5); let n = arr.length; document.write(findAnswer(n, arr)); // This code is contributed by _saurabh_jaiswal |
28465
Time Complexity: O(nlogn), used for sorting the array
Auxiliary Space: O(1), as no extra space is used