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

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

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

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. Prin 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



Output:

4 5 9

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