Intercambiar pares de límites alternativos

Dada una array arr[] de N enteros, la tarea es intercambiar el primer y el último elemento, luego el tercero y el antepenúltimo elemento, luego el quinto y el quinto y así sucesivamente. Imprima la array final después de todas las operaciones válidas.
 

Entrada: arr[] = {1, 2, 3, 4, 5, 6} 
Salida: 6 2 4 3 5 1 
Operación 1: Intercambiar 1 y 6 
Operación 2: Intercambiar 3 y 4
Entrada: arr[] = {5, 54, 12, 63, 45} 
Salida: 45 54 12 63 5 
 

Enfoque: inicialice el puntero i = 0 y j = N – 1, luego intercambie los elementos en estos punteros y actualice i = i + 2 y j = j – 2 . Repita estos pasos mientras i < j . Finalmente imprima la array actualizada.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to print
// the contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Function to update the array
void UpdateArr(int arr[], int n)
{
 
    // Initialize the pointers
    int i = 0, j = n - 1;
 
    // While there are elements to swap
    while (i < j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
 
        // Update the pointers
        i += 2;
        j -= 2;
    }
 
    // Print the updated array
    printArr(arr, n);
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    UpdateArr(arr, n);
 
    return 0;
}

// Java implementation of the above approach
class GFG
{
     
    // Utility function to print
    // the contents of an array
    static void printArr(int arr[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
     
    // Function to update the array
    static void UpdateArr(int arr[], int n)
    {
     
        // Initialize the pointers
        int i = 0, j = n - 1;
     
        // While there are elements to swap
        while (i < j)
        {
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
     
            // Update the pointers
            i += 2;
            j -= 2;
        }
     
        // Print the updated array
        printArr(arr, n);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 1, 2, 3, 4, 5, 6 };
        int n = arr.length;
     
        UpdateArr(arr, n);
    }
}
 
// This code is contributed by AnkitRai01

# Python3 implementation of the approach
 
# Utility function to print
# the contents of an array
def printArr(arr, n):
 
    for i in range(n):
        print(arr[i], end = " ");
 
# Function to update the array
def UpdateArr(arr, n):
 
    # Initialize the pointers
    i = 0;
    j = n - 1;
 
    # While there are elements to swap
    while (i < j):
        temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
 
        # Update the pointers
        i += 2;
        j -= 2;
     
    # Print the updated array
    printArr(arr, n);
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4, 5, 6 ];
    n = len(arr);
 
    UpdateArr(arr, n);
 
# This code is contributed by PrinciRaj1992

// C# implementation of the approach
using System;
     
class GFG
{
     
    // Utility function to print
    // the contents of an array
    static void printArr(int []arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
     
    // Function to update the array
    static void UpdateArr(int []arr, int n)
    {
     
        // Initialize the pointers
        int i = 0, j = n - 1;
     
        // While there are elements to swap
        while (i < j)
        {
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
     
            // Update the pointers
            i += 2;
            j -= 2;
        }
     
        // Print the updated array
        printArr(arr, n);
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int []arr = { 1, 2, 3, 4, 5, 6 };
        int n = arr.Length;
     
        UpdateArr(arr, n);
    }
}
 
// This code is contributed by 29AjayKumar

<script>
 
// JavaScript implementation of the approach
 
// Utility function to print
// the contents of an array
function printArr(arr, n)
{
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}
 
// Function to update the array
function UpdateArr(arr, n)
{
 
    // Initialize the pointers
    let i = 0, j = n - 1;
 
    // While there are elements to swap
    while (i < j) {
        let temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
 
        // Update the pointers
        i += 2;
        j -= 2;
    }
 
    // Print the updated array
    printArr(arr, n);
}
 
// Driver code
 
    let arr = [ 1, 2, 3, 4, 5, 6 ];
    let n = arr.length;;
 
    UpdateArr(arr, n);
 
// This code is contributed by Surbhi Tyagi
 
</script>

6 2 4 3 5 1

Complejidad temporal: O(n)
Espacio auxiliar: O(1)