Maximize sum of K pairs made up of elements that are equidistant from both ends of the array

Given an array arr[] consisting of N integers and an integer K, the task is to find the maximum sum of K pairs of the form (arr[i], arr[N - i - 1]), where (0 ? i ? N - 1).

Examples:

Input: arr[] = {2, -4, 3, -1, 2, 5}, K = 2
Output: 9
Explanation: All possibles pair of the form (arr[i], arr[N - i + 1]) are:

1. (2, 5)
2. (-1, 3)
3. (-4, 2)

From the above pairs, the K(= 2) pairs with maximum sum are (2, 5) and (-1, 3). Therefore, maximum sum = 2 + 5 + (-1) + 3 = 9.

Input: arr[] = {2, -4, -2, 4}, K = 2
Output: 0

Naive Approach: The simplest approach to solve the problem is to generate all possible pairs of the given form from the array and calculate the maximum sum of K such pairs. After checking for all the pairs, print the maximum sum obtained among all the possible sums.

 Time Complexity: O(N² * K)
Auxiliary Space: O(N²)

Efficient Approach: The above approach can be optimized greedily. The idea is to choose only the K pairs with highest cost. Follow the steps below to solve the problem:

• Initialize a vector, say pairwiseSum[], to store the sums of required pairs from the array.
• Generate pairs (arr[i], arr[N - i + 1]) from the given array by traversing the array. Store their sums in the vector pairwiseSum[].
• Sort the vector pairwiseSum[] in descending order.
• After completing the above steps, print the sum of the first K elements of the vector pairwiseSum[] as the resultant sum.

Below is the implementation of the above approach:

// C++ program for the above approach

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

// Function to find the maximum sum of
// K pairs of the form (arr[i],
// arr[N - i - 1])
int maxSum(int arr[], int N, int K)
{
    // Stores the resultant pairwise
    // sum
    vector<int> pairwiseSum;

    // Traverse till half of the array
    for (int i = 0; i < N / 2; i++) {

        // Update the value of curSum
        int curSum = arr[i] + arr[i + N / 2];
        pairwiseSum.push_back(curSum);
    }

    // Sort in the descending order
    sort(pairwiseSum.begin(),
         pairwiseSum.end(),
         greater<int>());

    // Stores the resultant maximum sum
    int maxSum = 0;

    // Add K maximum sums obtained
    for (int i = 0; i < K; i++) {

        // Update the value of maxSum
        maxSum += pairwiseSum[i];
    }

    // Print the maximum sum
    cout << maxSum;
}

// Driver Code
int main()
{
    int arr[] = { 2, -4, 3, 5, 2, -1 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
    maxSum(arr, N, K);

    return 0;
}

// Java program for the above approach
import java.util.ArrayList;
import java.util.Collections;
class GFG{

// Function to find the maximum sum of
// K pairs of the form (arr[i],
// arr[N - i - 1])
public static void maxSum(int arr[], int N, int K)
{
    // Stores the resultant pairwise
    // sum
    ArrayList<Integer> pairwiseSum = new ArrayList<Integer>();

    // Traverse till half of the array
    for (int i = 0; i < N / 2; i++) {

        // Update the value of curSum
        int curSum = arr[i] + arr[i + N / 2];
        pairwiseSum.add(curSum);
    }

    // Sort in the descending order
    Collections.sort(pairwiseSum);
    Collections.reverse(pairwiseSum);

    // Stores the resultant maximum sum
    int maxSum = 0;

    // Add K maximum sums obtained
    for (int i = 0; i < K; i++) {

        // Update the value of maxSum
        maxSum += pairwiseSum.get(i);
    }

    // Print the maximum sum
    System.out.println(maxSum);
}

// Driver Code
public static void main(String args[])
{
    int arr[] = { 2, -4, 3, 5, 2, -1 };
    int K = 2;
    int N = arr.length;
    maxSum(arr, N, K);
}
}

// This code is contributed by gfgking.

# Python3 program for the above approach

# Function to find the maximum sum of
# K pairs of the form (arr[i],
# arr[N - i - 1])
def maxSum(arr, N, K):
    
    # Stores the resultant pairwise
    # sum
    pairwiseSum = []

    # Traverse till half of the array
    for i in range(N // 2):
        
        # Update the value of curSum
        curSum = arr[i] + arr[i + N // 2]
        pairwiseSum.append(curSum)

    # Sort in the descending order
    pairwiseSum.sort(reverse = True)

    # Stores the resultant maximum sum
    maxSum = 0

    # Add K maximum sums obtained
    for i in range(K):
        
        # Update the value of maxSum
        maxSum += pairwiseSum[i]

    # Print the maximum sum
    print(maxSum)

# Driver Code
if __name__ == '__main__':
    
    arr = [ 2, -4, 3, 5, 2, -1 ]
    K = 2
    N = len(arr)
    
    maxSum(arr, N, K)

# This code is contributed by bgangwar59

// C# program for the above approach

using System;
using System.Collections.Generic;

class GFG{
 
// Function to find the maximum sum of
// K pairs of the form (arr[i],
// arr[N - i - 1])
static void maxSum(int []arr, int N, int K)
{
    // Stores the resultant pairwise
    // sum
    List<int> pairwiseSum = new List<int>();

    // Traverse till half of the array
    for (int i = 0; i < N / 2; i++) {

        // Update the value of curSum
        int curSum = arr[i] + arr[i + N / 2];
        pairwiseSum.Add(curSum);
    }

    // Sort in the descending order
    pairwiseSum.Sort();
    pairwiseSum.Reverse();
    // Stores the resultant maximum sum
    int maxSum = 0;

    // Add K maximum sums obtained
    for (int i = 0; i < K; i++) {

        // Update the value of maxSum
        maxSum += pairwiseSum[i];
    }

    // Print the maximum sum
    Console.Write(maxSum);
}

// Driver Code
public static void Main()
{
    int []arr = { 2, -4, 3, 5, 2, -1 };
    int K = 2;
    int N = arr.Length;
    maxSum(arr, N, K);

}
}

// This code is contributed by ipg2016107.

  <script>
  
        // JavaScript program for the above approach


        // Function to find the maximum sum of
        // K pairs of the form (arr[i],
        // arr[N - i - 1])
        function maxSum(arr, N, K) {
            // Stores the resultant pairwise
            // sum
            let pairwiseSum = [];

            // Traverse till half of the array
            for (let i = 0; i < N / 2; i++) {

                // Update the value of curSum
                let curSum = arr[i] + arr[i + parseInt(N / 2)];
                pairwiseSum.push(curSum);
            }

            // Sort in the descending order
            pairwiseSum.sort(function (a, b) { return b - a });

            // Stores the resultant maximum sum
           let maxSum = 0;

            // Add K maximum sums obtained
            for (let i = 0; i < K; i++) {

                // Update the value of maxSum
                maxSum += pairwiseSum[i];
            }

            // Print the maximum sum
            document.write(maxSum);
        }

        // Driver Code

        let arr = [2, -4, 3, 5, 2, -1];
        let K = 2;
        let N = arr.length;
        maxSum(arr, N, K);

  // This code is contributed by Potta Lokesh
  
  </script>

9

Time Complexity: O(N * log N)
Auxiliary Space: O(N)