Given an array arr[] of N integers where N % 4 = 0, the task is to divide the integers into groups of four such that when the combined sum of the maximum two elements from all the groups is taken, it is minimum possible. Print the minimized sum.
Examples:
Input: arr[] = {1, 1, 2, 2}
Output: 4
The only group will be {1, 1, 2, 2}.
2 + 2 = 4
Input: arr[] = {1, 1, 10, 2, 2, 2, 1, 8}
Output: 21
{1, 1, 2, 1} and {10, 2, 2, 8} are the groups that will
give the minimum sum as 1 + 2 + 10 + 8 = 21.
Approach: In order to minimize the sum, the maximum four elements from the array must be in the same group because the maximum two elements will definitely be included in the sum no matter what group they are a part of but the next two maximum elements can be prevented if they are part of this group. Making groups, in the same manner, will give the minimum sum possible. So, sort the array in descending order and starting from the first element, make groups of four consecutive elements.
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 minimum required sumint minSum(int arr[], int n){ // To store the required sum int sum = 0; // Sort the array in descending order sort(arr, arr + n, greater<int>()); for (int i = 0; i < n; i++) { // The indices which give 0 or 1 as // the remainder when divided by 4 // will be the maximum two // elements of the group if (i % 4 < 2) sum = sum + arr[i]; } return sum;}// Driver codeint main(){ int arr[] = { 1, 1, 10, 2, 2, 2, 1 }; int n = sizeof(arr) / sizeof(arr[0]); cout << minSum(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {// Function to return the minimum required sumstatic int minSum(Integer arr[], int n){ // To store the required sum int sum = 0; // Sort the array in descending order Arrays.sort(arr, Collections.reverseOrder()); for (int i = 0; i < n; i++) { // The indices which give 0 or 1 as // the remainder when divided by 4 // will be the maximum two // elements of the group if (i % 4 < 2) sum = sum + arr[i]; } return sum;}// Driver codepublic static void main(String[] args){ Integer []arr = { 1, 1, 10, 2, 2, 2, 1 }; int n = arr.length; System.out.println(minSum(arr, n));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # Function to return the minimum required sum def minSum(arr, n) : # To store the required sum sum = 0; # Sort the array in descending order arr.sort(reverse = True) for i in range(n) : # The indices which give 0 or 1 as # the remainder when divided by 4 # will be the maximum two # elements of the group if (i % 4 < 2) : sum += arr[i]; return sum; # Driver code if __name__ == "__main__" : arr = [ 1, 1, 10, 2, 2, 2, 1 ]; n = len(arr); print(minSum(arr, n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG {// Function to return the minimum required sumstatic int minSum(int []arr, int n){ // To store the required sum int sum = 0; // Sort the array in descending order Array.Sort(arr); Array.Reverse(arr); for (int i = 0; i < n; i++) { // The indices which give 0 or 1 as // the remainder when divided by 4 // will be the maximum two // elements of the group if (i % 4 < 2) sum = sum + arr[i]; } return sum;}// Driver codepublic static void Main(String[] args){ int []arr = { 1, 1, 10, 2, 2, 2, 1 }; int n = arr.Length; Console.WriteLine(minSum(arr, n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to return the minimum required sumfunction minSum(arr, n){ // To store the required sum let sum = 0; // Sort the array in descending order arr.sort(function(a, b){return b-a}); for (let i = 0; i < n; i++) { // The indices which give 0 or 1 as // the remainder when divided by 4 // will be the maximum two // elements of the group if (i % 4 < 2) sum = sum + arr[i]; } return sum;}// Driver code let arr = [ 1, 1, 10, 2, 2, 2, 1 ]; let n = arr.length; document.write(minSum(arr, n));</script> |
14
Time Complexity: O(n * log n)
Auxiliary Space: O(1)
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