Suma máxima de mínimos de pares en una array

Dada una array arr[] de N enteros donde N es par, la tarea es agrupar los elementos de la array en los pares (X1, Y1), (X2, Y2), (X3, Y3), … tal que la suma min( X1, Y1) + min(X2, Y2) + min(X3, Y3) + … es máximo.
Ejemplos: 
 

Entrada: arr[] = {1, 5, 3, 2} 
Salida: 4 
(1, 5) y (3, 2) -> 1 + 2 = 3 
(1, 3) y (5, 2) -> 1 + 2 = 3 
(1, 2) y (5, 3) -> 1 + 3 = 4
Entrada: arr[] = {1, 3, 2, 1, 4, 5} 
Salida: 7 
 

Enfoque: no importa cómo se formen los pares, el elemento máximo de la array siempre se ignorará, ya que será el elemento máximo en cada par en el que se coloque. Lo mismo ocurre con el segundo elemento máximo a menos que esté emparejado con el elemento máximo. Entonces, para maximizar la suma, un enfoque óptimo será ordenar la array y comenzar a hacer pares en orden a partir del elemento máximo.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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

Complejidad de Tiempo: O(n logn)
Espacio Auxiliar: O(1)