# 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:

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

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## Javascript

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Output:

`4 5 9`

Time Complexity: O(n)

Auxiliary Space: O(1)

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