Swap Alternate Boundary Pairs

Given an array arr[] of N integers, the task is to swap the first and the last element then the third and the third last element then fifth and fifth last and so on. Print the final array after all the valid operations.

Input: arr[] = {1, 2, 3, 4, 5, 6} 
Output: 6 2 4 3 5 1 
Operation 1: Swap 1 and 6 
Operation 2: Swap 3 and 4
Input: arr[] = {5, 54, 12, 63, 45} 
Output: 45 54 12 63 5 

Approach: Initialize pointer i = 0 and j = N – 1 then swap the elements at these pointers and update i = i + 2 and j = j – 2. Repeat these steps while i < j. Finally print the updated array.
Below is the implementation of the above approach: 

C++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

6 2 4 3 5 1

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