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

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++
Java
Python3
C#
PHP
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Utility function to print// the contents of an arrayvoid 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 1void 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 codeint main(){    int arr[] = { 3, 6, 8 };    int n = sizeof(arr) / sizeof(arr[0]);    updateArr(arr, n);     return 0;}
// Java implementation of the approachclass GfG{ // Utility function to print// the contents of an arraystatic 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 1static 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 codepublic 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 implementation of the approach # Utility function to print# the contents of an arraydef 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 1def 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 codeif __name__ == '__main__':    arr = [3, 6, 8];    n = len(arr);    updateArr(arr, n); # This code contributed by PrinciRaj1992
// C# implementation of the approachclass GfG{ // Utility function to print// the contents of an arraystatic 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 1static 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 codestatic void Main(){    int []arr = { 3, 6, 8 };    int n = arr.Length;    updateArr(arr, n); }} // This code is contributed by mits
<?php// PHP implementation of the approach // Utility function to print// the contents of an arrayfunction 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 1function 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?>
<script> // javascript implementation of the approach // Utility function to print// the contents of an arrayfunction printArr(arr, n){     var i;    for (i = 0; i < n; i++)      document.write(arr[i] + " ");} // Function to increment all the odd// positioned elements by 1 and decrement// all the even positioned elements by 1function updateArr(arr, n){    var i;    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    var arr = [3, 6, 8];    var n = arr.length;    updateArr(arr, n);     </script>
4 5 9
 
Time Complexity: O(n)
Auxiliary Space: O(1)
prerna saini
Mithun Kumar
princiraj1992
subham348
bgangwar59
anikakapoor
school-programming
Algorithms
Arrays
Arrays
Algorithms
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