Given an array arr[] of length N, the task is to select a quadruple (i, j, k, l) and calculate the sum of the second minimums of all possible quadruples.
Note: It is guaranteed that N is a multiple of 4 and each array element can be part of a single quadruple.
Examples:
Input: arr[] = {7, 4, 5, 2, 3, 1, 5, 9}
Output: 8
Explanation:
Quadruple 1: {7, 1, 5, 9} => 2nd Minimum value = 5.
Quadruple 2: {4, 5, 2, 3} => 2nd Minimum value = 3.
Therefore, the maximum possible sum is 8.Input: arr[] = {7, 4, 3, 3}
Output: 3
Approach: The idea is to use Greedy Approach to solve this problem. Below are the steps:
- Initialize a variable, say sum, to store the maximum gain from the array.
- Sort the array in ascending order.
- Traverse the array in reverse and add every third element encountered.
- Print the sum obtained.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <iostream>#include <bits/stdc++.h>using namespace std;// Function to find maximum possible sum of// second minimums in each quadruplevoid maxPossibleSum(int arr[], int N){ // Sort the array sort(arr, arr + N); int sum = 0; int j = N - 3; while (j >= 0) { // Add the second minimum sum += arr[j]; j -= 3; } // Print maximum possible sum cout << sum;}// Driver Codeint main(){ // Given array int arr[] = { 7, 4, 5, 2, 3, 1, 5, 9 }; // Size of the array int N = 8; maxPossibleSum(arr, N); return 0;}// This code is contributed by aditya7409 |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG { // Function to find maximum possible sum of // second minimums in each quadruple public static void maxPossibleSum(int[] arr, int N) { // Sort the array Arrays.sort(arr); int sum = 0; int j = N - 3; while (j >= 0) { // Add the second minimum sum += arr[j]; j -= 3; } // Print maximum possible sum System.out.println(sum); } // Driver Code public static void main(String[] args) { // Given array int[] arr = { 7, 4, 5, 2, 3, 1, 5, 9 }; // Size of the array int N = arr.length; maxPossibleSum(arr, N); }} |
Python3
# Python 3 program for the above approach# Function to find maximum possible sum of# second minimums in each quadrupledef maxPossibleSum(arr, N): # Sort the array arr.sort() sum = 0 j = N - 3 while (j >= 0): # Add the second minimum sum += arr[j] j -= 3 # Print maximum possible sum print(sum)# Driver Codeif __name__ == "__main__": # Given array arr = [7, 4, 5, 2, 3, 1, 5, 9] # Size of the array N = 8 maxPossibleSum(arr, N) # This code is contributed by chitranayal |
C#
// C# program for the above approachusing System;public class GFG{ // Function to find maximum possible sum of // second minimums in each quadruple public static void maxPossibleSum(int[] arr, int N) { // Sort the array Array.Sort(arr); int sum = 0; int j = N - 3; while (j >= 0) { // Add the second minimum sum += arr[j]; j -= 3; } // Print maximum possible sum Console.WriteLine(sum); } // Driver Code public static void Main(String[] args) { // Given array int[] arr = { 7, 4, 5, 2, 3, 1, 5, 9 }; // Size of the array int N = arr.Length; maxPossibleSum(arr, N); }}// This code is contributed by shikhasingrajput |
Javascript
<script>// javascript program of the above approach // Function to find maximum possible sum of // second minimums in each quadruple function maxPossibleSum(arr, N) { // Sort the array arr.sort(); let sum = 0; let j = N - 3; while (j >= 0) { // Add the second minimum sum += arr[j]; j -= 3; } // Print maximum possible sum document.write(sum); } // Driver Code // Given array let arr = [ 7, 4, 5, 2, 3, 1, 5, 9 ]; // Size of the array let N = arr.length; maxPossibleSum(arr, N);// Thiscode is contributed by target_2.</script> |
Output:
8
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
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