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:

#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;
}

#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;
}

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);
    }
}

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)

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);
    }
}

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);

Number of vowels: 3

Time Complexity: O(N), where N is the length of the input word.
Auxiliary Space: O(1)