How to Count Vowels in a String in Java

The Challenge

Count the number of vowels (a, e, i, o, u) in a string!

Solution

Method 1: Using Loop and contains()

java

String text = "Hello World";
int vowelCount = 0;
String vowels = "aeiouAEIOU";

for (int i = 0; i < text.length(); i++) {
    if (vowels.indexOf(text.charAt(i)) != -1) {
        vowelCount++;
    }
}

System.out.println("Vowel count: " + vowelCount);
// Output: Vowel count: 3 (e, o, o)

Method 2: Using Regex

java

String text = "Hello World";
int vowelCount = text.replaceAll("[^aeiouAEIOU]", "").length();

System.out.println("Vowel count: " + vowelCount);
// Output: Vowel count: 3

Method 3: Using Switch Statement

java

String text = "Hello World";
int vowelCount = 0;

for (char c : text.toLowerCase().toCharArray()) {
    switch (c) {
        case 'a':
        case 'e':
        case 'i':
        case 'o':
        case 'u':
            vowelCount++;
            break;
    }
}

System.out.println("Vowel count: " + vowelCount);
// Output: Vowel count: 3

Did You Know?

The letter 'y' can sometimes act as a vowel (like in "gym" or "fly"), but it's traditionally classified as a consonant in most programming challenges!

Pro Tip

Use Java 8 streams for a functional approach:

java

long vowelCount = text.toLowerCase()
                    .chars()
                    .filter(c -> "aeiou".indexOf(c) != -1)
                    .count();

Real-World Use Cases

Linguistics: Analyze text patterns
Password strength: Check character diversity
Text games: Word puzzles and challenges
Data validation: Verify input format

›Short versionOpen if you only have a minute.

Walk the chars and test membership in a vowel set.
A lowercase-only set misses A, E, I, O, and U.
toLowerCase() without a locale follows the default locale. Turkish maps I to ı, not i.
y is a vowel only if this recipe says so. The page does not count it.

Count vowels

Count characters that belong to the vowel set a, e, i, o, u, ignoring case.

Working approach

Fold case with toLowerCase(Locale.ROOT) on each character, or on a copy of the string.
Test "aeiou".indexOf(c) >= 0.
Leave y, digits, and punctuation uncounted.

Walk the sample

Input "Java"
J is not a vowel
Both a characters match
The count is 2

The trap

String.toLowerCase() with no locale uses the JVM default. On a Turkish locale, "I" does not become "i", and an English vowel count drops those letters.

java

static int vowelCount(String text) {
    int count = 0;
    for (int i = 0; i < text.length(); i++) {
        char c = Character.toLowerCase(text.charAt(i));
        if ("aeiou".indexOf(c) >= 0) {
            count++;
        }
    }
    return count;
}

Why can "AEIOU" report a vowel count of 0?

The set "aeiou" does not contain A. indexOf looks for the same char.

Fold case first. Prefer Character.toLowerCase(char), which is not the Turkish string rule, or String.toLowerCase(Locale.ROOT) if you lowercase the whole text.

Time O(n), space O(1).

Easy to get wrong

Comparing only against lowercase literals.
Calling toLowerCase() with no locale on text that includes I.