# Increment odd positioned elements by 1 and decrement even positioned elements by 1 in an Array

• Difficulty Level : Easy
• Last Updated : 29 Sep, 2021

Given an array arr[], the task is increment all the odd positioned elements by 1 and decrement all the even positioned elements by 1.

Examples:

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[] = {3, 6, 8}
Output: 4 5 9

Input: arr[] = {9, 7, 3}
Output: 10 6 4

Approach: Traverse the array element by element and if the current element’s position is odd then increment it by 1 else decrement it by 1. Print the contents of the updated array in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``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 increment all the odd``// positioned elements by 1 and decrement``// all the even positioned elements by 1``void` `updateArr(``int` `arr[], ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++)` `        ``// If current element is odd positioned``        ``if` `((i + 1) % 2 == 1)``            ``arr[i]++;` `        ``// If even positioned``        ``else``            ``arr[i]--;` `    ``// Print the updated array``    ``printArr(arr, n);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 3, 6, 8 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``updateArr(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the 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 increment all the odd``// positioned elements by 1 and decrement``// all the even positioned elements by 1``static` `void` `updateArr(``int` `arr[], ``int` `n)``{``    ``for` `(``int` `i = ``0``; i < n; i++)` `        ``// If current element is odd positioned``        ``if` `((i + ``1``) % ``2` `== ``1``)``            ``arr[i]++;` `        ``// If even positioned``        ``else``            ``arr[i]--;` `    ``// Print the updated array``    ``printArr(arr, n);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``3``, ``6``, ``8` `};``    ``int` `n = arr.length;``    ``updateArr(arr, n);` `}``}` `// This code is contributed by Prerna Saini`

## Python3

 `# Python3 implementation of the approach` `# Utility function to print``# the contents of an array``def` `printArr(arr, n):``    ``for` `i ``in` `range``(``0``, n):``        ``print``(arr[i], end ``=` `" "``);` `# Function to increment all the odd``# positioned elements by 1 and decrement``# all the even positioned elements by 1``def` `updateArr(arr, n):``    ``for` `i ``in` `range``(``0``, n):` `        ``# If current element is odd positioned``        ``if` `((i ``+` `1``) ``%` `2` `=``=` `1``):``            ``arr[i] ``+``=` `1``;` `        ``# If even positioned``        ``else``:``            ``arr[i] ``-``=` `1``;` `    ``# Print the updated array``    ``printArr(arr, n);` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``3``, ``6``, ``8``];``    ``n ``=` `len``(arr);``    ``updateArr(arr, n);` `# This code contributed by PrinciRaj1992`

## C#

 `// C# implementation of the 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.Console.Write(arr[i] + ``" "``);``}` `// Function to increment all the odd``// positioned elements by 1 and decrement``// all the even positioned elements by 1``static` `void` `updateArr(``int` `[]arr, ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++)` `        ``// If current element is odd positioned``        ``if` `((i + 1) % 2 == 1)``            ``arr[i]++;` `        ``// If even positioned``        ``else``            ``arr[i]--;` `    ``// Print the updated array``    ``printArr(arr, n);``}` `// Driver code``static` `void` `Main()``{``    ``int` `[]arr = { 3, 6, 8 };``    ``int` `n = arr.Length;``    ``updateArr(arr, n);` `}``}` `// This code is contributed by mits`

## PHP

 ``

## Javascript

 ``
Output:
`4 5 9`

Time Complexity: O(n)

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up