Maximum sum of minimums of pairs in an array

Given an array arr[] of N integers where N is even, the task is to group the array elements in the pairs (X1, Y1), (X2, Y2), (X3, Y3), ... such that the sum min(X1, Y1) + min(X2, Y2) + min(X3, Y3) + ... is maximum.
Examples: 
 

Input: arr[] = {1, 5, 3, 2} 
Output: 4 
(1, 5) and (3, 2) -> 1 + 2 = 3 
(1, 3) and (5, 2) -> 1 + 2 = 3 
(1, 2) and (5, 3) -> 1 + 3 = 4
Input: arr[] = {1, 3, 2, 1, 4, 5} 
Output: 7 
 

Approach: No matter how the pairs are formed, the maximum element from the array will always be ignored as it will be the maximum element in every pair it is put into. Same goes for the second maximum element unless it is paired with the maximum element. So, to maximize the sum an optimal approach will be to sort the array and start making pairs in order starting from the maximum element.
Below is the implementation of the above approach: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the maximum
// required sum of the pairs
int maxSum(int a[], int n)
{

    // Sort the array
    sort(a, a + n);

    // To store the sum
    int sum = 0;

    // Start making pairs of every two
    // consecutive elements as n is even
    for (int i = 0; i < n - 1; i += 2) {

        // Minimum element of the current pair
        sum += a[i];
    }

    // Return the maximum possible sum
    return sum;
}

// Driver code
int main()
{
    int arr[] = { 1, 3, 2, 1, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << maxSum(arr, n);

    return 0;
}

// Java implementation of the approach
import java.util.Arrays;

class GFG
{
    
// Function to return the maximum
// required sum of the pairs
static int maxSum(int a[], int n)
{

    // Sort the array
    Arrays.sort(a);

    // To store the sum
    int sum = 0;

    // Start making pairs of every two
    // consecutive elements as n is even
    for (int i = 0; i < n - 1; i += 2)
    {

        // Minimum element of the current pair
        sum += a[i];
    }

    // Return the maximum possible sum
    return sum;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 3, 2, 1, 4, 5 };
    int n = arr.length;

    System.out.println(maxSum(arr, n));
}
}

// This code is contributed by Code_Mech

# Python3 implementation of the approach 

# Function to return the maximum 
# required sum of the pairs 
def maxSum(a, n) :

    # Sort the array 
    a.sort(); 

    # To store the sum 
    sum = 0; 

    # Start making pairs of every two 
    # consecutive elements as n is even 
    for i in range(0, n - 1, 2) : 

        # Minimum element of the current pair 
        sum += a[i]; 

    # Return the maximum possible sum 
    return sum; 

# Driver code 
if __name__ == "__main__" : 

    arr = [ 1, 3, 2, 1, 4, 5 ]; 
    n = len(arr); 

    print(maxSum(arr, n));

# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
    
class GFG
{
    
// Function to return the maximum
// required sum of the pairs
static int maxSum(int []a, int n)
{

    // Sort the array
    Array.Sort(a);

    // To store the sum
    int sum = 0;

    // Start making pairs of every two
    // consecutive elements as n is even
    for (int i = 0; i < n - 1; i += 2)
    {

        // Minimum element of the current pair
        sum += a[i];
    }

    // Return the maximum possible sum
    return sum;
}

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

    Console.WriteLine(maxSum(arr, n));
}
}

// This code is contributed by 29AjayKumar

<script>
// Javascript implementation of the approach

// Function to return the maximum
// required sum of the pairs
function maxSum(a, n) {

    // Sort the array
    a.sort((a, b) => a - b);

    // To store the sum
    let sum = 0;

    // Start making pairs of every two
    // consecutive elements as n is even
    for (let i = 0; i < n - 1; i += 2) {

        // Minimum element of the current pair
        sum += a[i];
    }

    // Return the maximum possible sum
    return sum;
}

// Driver code
let arr = [1, 3, 2, 1, 4, 5];
let n = arr.length;

document.write(maxSum(arr, n));


// This code is contributed by _saurabh_jaiswal
</script>

7

Time Complexity: O(n logn)
Auxiliary Space: O(1)