C-Sharp | Java | Python | Swift | GO | WPF | Ruby | Scala | F# | JavaScript | SQL | PHP | Angular | HTML
Q. Program to print the elements of an array present on odd position.ExplanationIn this program, we need to print the elements of the array which are present in odd positions. This can be accomplished by looping through the array and printing the elements of an array by incrementing i by 2 till the end of the array is reached. In the above array, the elements which are on odd positions are a, c and e. Algorithm
SolutionPython#Initialize array arr = [1, 2, 3, 4, 5]; print("Elements of given array present on odd position: "); #Loop through the array by incrementing the value of i by 2 for i in range(0, len(arr), 2): print(arr[i]); Output: Elements of given array present on odd position: 1 3 5 C#include <stdio.h> int main() { //Initialize array int arr[] = {1, 2, 3, 4, 5}; //Calculate length of array arr int length = sizeof(arr)/sizeof(arr[0]); printf("Elements of given array present on odd position: \n"); //Loop through the array by incrementing value of i by 2 for (int i = 0; i < length; i = i+2) { printf("%d\n", arr[i]); } return 0; } Output: Elements of given array present on odd position: 1 3 5 JAVApublic class OddPosition { public static void main(String[] args) { //Initialize array int [] arr = new int [] {1, 2, 3, 4, 5}; System.out.println("Elements of given array present on odd position: "); //Loop through the array by incrementing value of i by 2 for (int i = 0; i < arr.length; i = i+2) { System.out.println(arr[i]); } } } Output: Elements of given array present on odd position: 1 3 5 C#using System; public class OddPosition { public static void Main() { //Initialize array int [] arr = new int [] {1, 2, 3, 4, 5}; Console.WriteLine("Elements of given array present on odd position: "); //Loop through the array by incrementing value of i by 2 for (int i = 0; i < arr.Length; i = i+2) { Console.WriteLine(arr[i]); } } } Output: Elements of given array present on odd position: 1 3 5 PHP<!DOCTYPE html> <html> <body> <?php //Initialize array $arr = array(1, 2, 3, 4, 5); print("Elements of given array present on odd position: <br>"); //Loop through the array by incrementing value of i by 2 for ($i = 0; $i < count($arr); $i = $i+2) { print($arr[$i] . "<br>"); } ?> </body> </html> Output: Elements of given array present on odd position: 1 3 5
Next Topic#
|