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)