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

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 <bits/stdc++.h>
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

<?php
// PHP implementation of the approach

// Utility function to print
// the contents of an array
function printArr($arr, $n)
{
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i] . " ";
}

// Function to increment all the odd
// positioned elements by 1 and decrement
// all the even positioned elements by 1
function updateArr($arr, $n)
{
    for ($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
$arr = array( 3, 6, 8 );
$n = count($arr);
updateArr($arr, $n);

// This code is contributed by mits
?>

Output:

4 5 9

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