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

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  • Difficulty Level : Hard
  • Last Updated : 28 Jun, 2022
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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: 

  1. 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.
  2. 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: 
 

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#




// 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: 

10

 

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


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