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Count Vowels in a String in Java

beginner· Strings · String

Problem

Vowels are the letters a, e, i, o, and u — every other letter is a consonant.

Given a string, count how many vowels it contains.

Input

Programming

Output

Vowels: 3

Java Program

Java

public class CountVowels {
    public static void main(String[] args) {
        String str = "Programming";
        String vowelSet = "aeiouAEIOU";
        int count = 0;
        // Scan each character once, checking membership in the vowel set
        for (char c : str.toCharArray()) {
            if (vowelSet.indexOf(c) != -1) count++;
        }
        System.out.println("Vowels: " + count);
    }
}

▶ Run Code

Output

Vowels: 3

Core Logic

A single pass through the string, checking each letter against a small set of vowels, is all it takes.

How It Works

1str.toCharArray() converts the string into a char[] so it can be scanned one character at a time.
2vowelSet.indexOf(c) checks whether the current character appears in "aeiouAEIOU".
3A match increments the count variable; anything else is skipped.
4After the loop, count holds the total number of vowels found.

For "Programming", the scan finds 'o', 'a', and 'i' — three vowels in total.

💡

Key Point: Checking both cases in vowelSet avoids a separate call to toLowerCase() for every character.

Complexity

Time Complexity: O(n)Space Complexity: O(1)

Why: Each character is visited once, and only a single running counter is kept regardless of string length.

Key Concepts

toCharArray()String.indexOf()for-each loop

Approach 2: Regex Replace

Java

public class CountVowelsRegex {
    public static void main(String[] args) {
        String str = "Programming";
        // Removes every character that isn't a vowel, leaving only vowels behind
        String vowelsOnly = str.replaceAll("[^aeiouAEIOU]", "");
        System.out.println("Vowels: " + vowelsOnly.length());
    }
}

▶ Run Code

Output

Vowels: 3

Core Logic

A regex can strip out every non-vowel in one call, leaving only the vowels behind to count.

How It Works

1str.replaceAll("[^aeiouAEIOU]", "") matches every character that is NOT a vowel and removes it.
2What's left is a new string containing only the vowels from the original.
3.length() on that filtered string gives the total vowel count directly.

Replacing every non-vowel in "Programming" leaves "oai", whose length is 3.

💡

Key Point: Regex engines compile the pattern before scanning, which adds overhead a plain loop doesn't have — fine for one-off use, but a hot loop calling this repeatedly would do better with the manual scan.

Complexity

Time Complexity: O(n)Space Complexity: O(n)

Why: replaceAll() builds an entirely new filtered string before length() can read its size.

Key Concepts

regexreplaceAll()String.length()

Approach 3: Java 8

Java

public class CountVowelsStream {
    public static void main(String[] args) {
        String str = "Programming";
        String vowelSet = "aeiouAEIOU";
        // filter() keeps only vowel character codes; count() reduces to a single total
        long count = str.chars().filter(c -> vowelSet.indexOf(c) != -1).count();
        System.out.println("Vowels: " + count);
    }
}

▶ Run Code

Output

Vowels: 3

Core Logic

The same classification reads as a stream pipeline — filter for vowels, then count what's left.

How It Works

1str.chars() returns an IntStream of the string's character codes.
2.filter(c -> vowelSet.indexOf(c) != -1) keeps only the codes that appear in the vowel set.
3.count() reduces the filtered stream down to a single long total.

Filtering "Programming" keeps only 'o', 'a', and 'i', so count() returns 3.

💡

Key Point: count() on a filtered stream reads close to the English description of the problem — 'count the characters that are vowels' — at the cost of a small amount of stream overhead versus the plain loop.

Complexity

Time Complexity: O(n)Space Complexity: O(1)

Why: The stream still visits every character once, and count() reduces straight down to a single long without collecting anything.

Key Concepts

Streamchars()filter()count()

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