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 9Input: 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 <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
// 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 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#
// 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 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?> |
Javascript
<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> |
Output:
4 5 9
Time Complexity: O(n)
Auxiliary Space: O(1)
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