Maximize 3rd element sum in quadruplet sets formed from given Array

Last Updated : 12 Jul, 2025

Given an array arr containing N values describing the priority of N jobs. The task is to form sets of quadruplets (W, X, Y, Z) to be done each day such that W >= X >= Y >= Z and in doing so, maximize the sum of all Y across all quadruplet sets.
Note: N will always be a multiple of 4.
Examples:

Input: Arr[] = {2, 1, 7, 5, 5, 4, 1, 1, 3, 3, 2, 2}
Output: 10
Explanation:
We can form 3 quadruplet sets as [7, 5, 5, 1], [4, 3, 3, 1], [2, 2, 2, 1].
The summation of all Y's are 5 + 3 + 2 = 10 which is the maximum possible value.
Input: arr[] = {1, 51, 91, 1, 1, 16, 1, 51, 48, 16, 1, 49}
Output: 68

Approach: To solve the problem mentioned above we can observe that:

Irrespective of Y, (W, X) >= Y, i.e., higher values of W and X are always lost and don't contribute to the answer. Therefore, we must keep these values as low as possible but greater or equal to Y.
Similarly, value for Z is always lost and must be less than Y. Therefore, it must be as low as possible.

Hence, to satisfy the above condition we have to:

first sort given array in descending order.
initialize a pointer that points to the third element of each pair from index 0.
subtract count of such pairs from the size of array i.e, N.

Below is the implementation of the above approach:

C++

// C++ code to Maximize 3rd element
// sum in quadruplet sets formed
// from given Array

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum
// possible value of Y
int formQuadruplets(int arr[], int n)
{

    int ans = 0, pairs = 0;

    // pairs contain count
    // of minimum elements
    // that will be utilized
    // at place of Z.
    // it is equal to count of
    // possible pairs that
    // is size of array divided by 4
    pairs = n / 4;

    // sorting the array in descending order
    // so as to bring values with minimal
    // difference closer to arr[i]
    sort(arr, arr + n, greater<int>());

    for (int i = 0; i < n - pairs; i += 3) {

        // here, i+2 acts as a
        // pointer that points
        // to the third value of
        // every possible quadruplet
        ans += arr[i + 2];
    }

    // returning the optimally
    // maximum possible value
    return ans;
}

// Driver code
int main()
{
    // array declaration
    int arr[] = { 2, 1, 7, 5, 5,
                  4, 1, 1, 3, 3,
                  2, 2 };

    // size of array
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << formQuadruplets(arr, n)
         << endl;

    return 0;
}

Java

// Java code to Maximize 3rd element
// sum in quadruplet sets formed
// from given Array
import java.util.*;
class GFG{

// Function to find the maximum
// possible value of Y
static int formQuadruplets(Integer arr[], int n)
{
    int ans = 0, pairs = 0;

    // pairs contain count
    // of minimum elements
    // that will be utilized
    // at place of Z.
    // it is equal to count of
    // possible pairs that
    // is size of array divided by 4
    pairs = n / 4;

    // sorting the array in descending order
    // so as to bring values with minimal
    // difference closer to arr[i]
    Arrays.sort(arr, Collections.reverseOrder());

    for (int i = 0; i < n - pairs; i += 3) 
    {

        // here, i+2 acts as a
        // pointer that points
        // to the third value of
        // every possible quadruplet
        ans += arr[i + 2];
    }

    // returning the optimally
    // maximum possible value
    return ans;
}

// Driver code
public static void main(String[] args)
{
    // array declaration
    Integer arr[] = { 2, 1, 7, 5, 5, 4, 
                      1, 1, 3, 3, 2, 2 };

    // size of array
    int n = arr.length;

    System.out.print(
           formQuadruplets(arr, n) + "\n");

}
}

// This code contributed by Rajput-Ji

Python3

# Python3 code to maximize 3rd element 
# sum in quadruplet sets formed 
# from given Array 

# Function to find the maximum 
# possible value of Y 
def formQuadruplets(arr, n): 

    ans = 0
    pairs = 0

    # Pairs contain count of minimum
    # elements that will be utilized 
    # at place of Z. It is equal to  
    # count of possible pairs that 
    # is size of array divided by 4 
    pairs = n // 4

    # Sorting the array in descending order 
    # so as to bring values with minimal 
    # difference closer to arr[i] 
    arr.sort(reverse = True)

    for i in range(0, n - pairs, 3): 

        # Here, i+2 acts as a pointer that  
        # points to the third value of 
        # every possible quadruplet 
        ans += arr[i + 2] 

    # Returning the optimally 
    # maximum possible value 
    return ans 

# Driver code 

# Array declaration 
arr = [ 2, 1, 7, 5, 5, 4, 1, 1, 3, 3, 2, 2 ] 

# Size of array 
n = len(arr) 

print(formQuadruplets(arr, n)) 

# This code is contributed by divyamohan123

// C# code to maximize 3rd element
// sum in quadruplet sets formed
// from given Array
using System;

class GFG{

// Function to find the maximum
// possible value of Y
static int formQuadruplets(int []arr, int n)
{
    int ans = 0, pairs = 0;

    // Pairs contain count of minimum  
    // elements that will be utilized at 
    // place of Z. It is equal to count of
    // possible pairs that is size of 
    // array divided by 4
    pairs = n / 4;

    // Sorting the array in descending order
    // so as to bring values with minimal
    // difference closer to arr[i]
    Array.Sort(arr);
    Array.Reverse(arr);
    for(int i = 0; i < n - pairs; i += 3) 
    {
       
       // Here, i+2 acts as a
       // pointer that points
       // to the third value of
       // every possible quadruplet
       ans += arr[i + 2];
    }

    // Returning the optimally
    // maximum possible value
    return ans;
}

// Driver code
public static void Main(String[] args)
{
    
    // Array declaration
    int []arr = { 2, 1, 7, 5, 5, 4, 
                  1, 1, 3, 3, 2, 2 };

    // Size of array
    int n = arr.Length;

    Console.Write(formQuadruplets(arr, n) + "\n");
}
}

// This code is contributed by amal kumar choubey

JavaScript

<script>

      // JavaScript code to maximize 3rd element
      // sum in quadruplet sets formed
      // from given Array
      
      // Function to find the maximum
      // possible value of Y
      function formQuadruplets(arr, n) {
        var ans = 0,
          pairs = 0;

        // Pairs contain count of minimum
        // elements that will be utilized at
        // place of Z. It is equal to count of
        // possible pairs that is size of
        // array divided by 4
        pairs = parseInt(n / 4);

        // Sorting the array in descending order
        // so as to bring values with minimal
        // difference closer to arr[i]
        arr.sort().reverse();
        for (var i = 0; i < n - pairs; i += 3) {
          // Here, i+2 acts as a
          // pointer that points
          // to the third value of
          // every possible quadruplet
          ans += arr[i + 2];
        }

        // Returning the optimally
        // maximum possible value
        return ans;
      }

      // Driver code
      // Array declaration
      var arr = [2, 1, 7, 5, 5, 4, 1, 1, 3, 3, 2, 2];

      // Size of array
      var n = arr.length;

      document.write(formQuadruplets(arr, n) + "<br>");
      
</script>

Output:

Time Complexity: O(n*log(n))
Auxiliary Space: O(1)

Analysis of Algorithms

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