Reverse the elements only at odd positions in the given Array

Given an array arr[] containing N integers, the task is to rearrange the array such that the odd indexed elements are in reverse order.

Examples:

Input: arr[] = {5, 7, 6, 2, 9, 18, 11, 15} 
Output: {5, 15, 6, 18, 9, 2, 11, 7} 
Explanation: 
The elements at even index [5, 6, 9, 11] are unchanged and elements at odd index are reversed from [7, 2, 18, 15] to [15, 18, 2, 7].

Input: arr[] = {1, 2, 3, 4, 5, 6} 
Output: {1, 6, 3, 4, 5, 2} 
Explanation: 
The elements at even index are unchanged and elements at odd index are reversed from [2, 4, 6] to [6, 4, 2]. 

Approach: To solve the problem mentioned above follow the steps given below: 

• Push the elements of odd indexes of the given array into a stack data structure.
• Replace the current array elements at odd indexes with elements at top of the stack and keep popping until the stack is empty. 
  

Below is the implementation of the above approach:

C++

// C++ program to reverse the
// elements only at odd positions
// in the given Array

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

// Function to display elements
void show(int arr[], int n)
{
    cout << "{";

    for (int i = 0; i < n - 1; i++)
        cout << arr[i] << ", ";

    cout << arr[n - 1] << "}";
}

// Function to flip elements
// at odd indexes
void flipHalf(int arr[], int n)
{
    int c = 0;
    int dup = n;

    stack<int> st;

    // Pushing elements at odd indexes
    // of a array to a stack
    for (int i = 0; i < n; i++) {
        int x = arr[i];

        if (c % 2 == 1)
            st.push(x);
        c++;
    }

    c = 0;

    // Replacing current elements at odd
    // indexes with element at top of stack
    for (int i = 0; i < n; i++) {
        int x = arr[i];

        if (c % 2 == 1) {
            x = st.top();
            st.pop();
        }
        arr[i] = x;
        c++;
    }
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };

    int n = sizeof(arr) / sizeof(arr[0]);

    flipHalf(arr, n);

    show(arr, n);

    return 0;
}

Java

// Java program to reverse the
// elements only at odd positions
// in the given array
import java.io.*;
import java.util.*;

class GFG {

// Function to count the valley points
// in the given character array
static void show(int arr[], int n)
{
    System.out.print("{");

    for(int i = 0; i < n - 1; i++)
       System.out.print(arr[i] + ", ");

    System.out.print(arr[n - 1] + "}");
}

// Function to flip elements
// at odd indexes
public static void flipHalf(int arr[], int n)
{
    int c = 0;
    int dup = n;
    Stack<Integer> st = new Stack<>();

    // Pushing elements at odd indexes
    // of a array to a stack
    for(int i = 0; i < n; i++)
    {
       int x = arr[i];

       if (c % 2 == 1)
       {
           st.push(x);
       }
       c++;
    }
    c = 0;

    // Replacing current elements at odd
    // indexes with element at top of stack
    for(int i = 0; i < n; i++)
    {
       int x = arr[i];

       if (c % 2 == 1)
       {
           x = st.peek();
           st.pop();
       }
       arr[i] = x;
       c++;
    }
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = arr.length;

    flipHalf(arr, n);
    show(arr, n);
}
}

// This code is contributed by Aman Kumar 27

Python3

# Python3 program to reverse the
# elements only at odd positions
# in the given Array

# Function to get top of the stack
def peek_stack(stack):

    if stack:
        return stack[-1]

# Function to display elements
def show(arr, n):

    print("{", end = " ")
    for i in range(0, n - 1):
        print(arr[i], ",", end = " ")

    print(arr[n - 1] , "}")

# Function to flip elements
# at odd indexes
def flipHalf(arr, n):

    c = 0
    dup = n
    stack = []

    # Pushing elements at odd indexes
    # of a array to a stack
    for i in range(0, n):
        x = arr[i]

        if c % 2 == 1:
            stack.append(x)

        c = c + 1

    c = 0

    # Replacing current elements at odd
    # indexes with element at top of stack
    for i in range(0, n):
        x = arr[i]

        if c % 2 == 1:
            x = peek_stack(stack)
            stack.pop()

        arr[i] = x
        c = c + 1

# Driver Code
if __name__ == "__main__":

  arr = [ 1, 2, 3, 4, 5, 6 ]
  n = len(arr)

  flipHalf(arr, n)

  show(arr, n)

# This code is contributed by akhilsaini

C#

// C# program to reverse the
// elements only at odd positions
// in the given array
using System;
using System.Collections.Generic;

class GFG{

// Function to count the valley points
// in the given character array
static void show(int []arr, int n)
{
    Console.Write("{");

    for(int i = 0; i < n - 1; i++)
    Console.Write(arr[i] + ", ");

    Console.Write(arr[n - 1] + "}");
}

// Function to flip elements
// at odd indexes
public static void flipHalf(int []arr, int n)
{
    int c = 0;
    int dup = n;
    Stack<int> st = new Stack<int>();

    // Pushing elements at odd indexes
    // of a array to a stack
    for(int i = 0; i < n; i++)
    {
        int x = arr[i];

        if (c % 2 == 1)
        {
            st.Push(x);
        }
        c++;
    }
    c = 0;

    // Replacing current elements at odd
    // indexes with element at top of stack
    for(int i = 0; i < n; i++)
    {
        int x = arr[i];

        if (c % 2 == 1)
        {
            x = st.Peek();
            st.Pop();
        }
        arr[i] = x;
        c++;
    }
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6 };
    int n = arr.Length;

    flipHalf(arr, n);
    show(arr, n);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
    // Javascript program to reverse the
    // elements only at odd positions
    // in the given Array

    // Function to display elements
    function show(arr, n)
    {
        document.write("{");

        for (let i = 0; i < n - 1; i++)
            document.write(arr[i] + ", ");

        document.write(arr[n - 1] + "}");
    }

    // Function to flip elements
    // at odd indexes
    function flipHalf(arr, n)
    {
        let c = 0;
        let dup = n;

        let st = [];

        // Pushing elements at odd indexes
        // of a array to a stack
        for (let i = 0; i < n; i++) {
            let x = arr[i];

            if (c % 2 == 1)
                st.push(x);
            c++;
        }

        c = 0;

        // Replacing current elements at odd
        // indexes with element at top of stack
        for (let i = 0; i < n; i++) {
            let x = arr[i];

            if (c % 2 == 1) {
                x = st[st.length - 1];
                st.pop();
            }
            arr[i] = x;
            c++;
        }
    }

    let arr = [ 1, 2, 3, 4, 5, 6 ];

    let n = arr.length;

    flipHalf(arr, n);

    show(arr, n);

</script>

Output:

{1, 6, 3, 4, 5, 2}

Time complexity: O(N)
Auxiliary Space Complexity: O(N/2)