# 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].

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 above approach:

## C++

 `// C++ program to reverse the ` `// elements only at odd positions ` `// in the given Array ` ` `  `#include ` `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); ` ` `  `    ``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 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 `

## 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 `

Output:

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

Time complexity: O(N)

Auxiliary Space Complexity: O(N/2)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : Aman Kumar 27, 29AjayKumar