Program to count the number of vowels in a word
Write a program to count the number of vowels in a given word. Vowels include ‘a’, ‘e’, ‘i’, ‘o’, and ‘u’ (both uppercase and lowercase). The program should output the count of vowels in the word.
Examples:
Input: “programming”
Output: 3
Explanation: The word “programming” has three vowels (‘o’, ‘a’, ‘i’).Input: “HELLO”
Output: 2
Explanation: The word “HELLO” has two vowels (‘E’, ‘O’).
Approach: To solve the problem, follow the below idea:
Iterate through each character in the given word and check if it is a vowel. If it is, increment the count.
Step-by-step algorithm:
- Initialize a variable
vowelCount
to 0 to keep track of the number of vowels. - Iterate through each character in the word.
- Check if the character is a vowel (either uppercase or lowercase).
- If it is a vowel, increment
vowelCount
. - After iterating through all characters,
vowelCount
will contain the total count of vowels.
Below is the implementation of the algorithm:
C++
#include <iostream> using namespace std; int main() { char word[] = "programming" ; int vowelCount = 0; for ( int i = 0; word[i] != '\0' ; ++i) { if (word[i] == 'a' || word[i] == 'e' || word[i] == 'i' || word[i] == 'o' || word[i] == 'u' || word[i] == 'A' || word[i] == 'E' || word[i] == 'I' || word[i] == 'O' || word[i] == 'U' ) { ++vowelCount; } } cout << "Number of vowels: " << vowelCount << endl; return 0; } |
C
#include <stdio.h> int main() { char word[] = "programming" ; int vowelCount = 0; for ( int i = 0; word[i] != '\0' ; ++i) { if (word[i] == 'a' || word[i] == 'e' || word[i] == 'i' || word[i] == 'o' || word[i] == 'u' || word[i] == 'A' || word[i] == 'E' || word[i] == 'I' || word[i] == 'O' || word[i] == 'U' ) { ++vowelCount; } } printf ( "Number of vowels: %d\n" , vowelCount); return 0; } |
Java
public class VowelCount { public static void main(String[] args) { String word = "programming" ; int vowelCount = 0 ; for ( int i = 0 ; i < word.length(); i++) { char currentChar = word.charAt(i); if (currentChar == 'a' || currentChar == 'e' || currentChar == 'i' || currentChar == 'o' || currentChar == 'u' || currentChar == 'A' || currentChar == 'E' || currentChar == 'I' || currentChar == 'O' || currentChar == 'U' ) { vowelCount++; } } System.out.println( "Number of vowels: " + vowelCount); } } |
Python3
word = "programming" vowel_count = 0 for char in word: if char.lower() in [ 'a' , 'e' , 'i' , 'o' , 'u' ]: vowel_count + = 1 print ( "Number of vowels:" , vowel_count) |
C#
using System; class Program { static void Main() { string word = "programming" ; int vowelCount = 0; foreach ( char c in word) { if ( "aeiouAEIOU" .Contains(c)) { vowelCount++; } } Console.WriteLine( "Number of vowels: " + vowelCount); } } |
Javascript
let word = "programming" ; let vowelCount = 0; for (let i = 0; i < word.length; i++) { let currentChar = word[i]; if ( 'aeiouAEIOU' .includes(currentChar)) { vowelCount++; } } console.log( "Number of vowels:" , vowelCount); |
Output
Number of vowels: 3
Time Complexity: O(N), where N is the length of the input word.
Auxiliary Space: O(1)