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Count Consonants in a String in Java

beginner· Strings · String

Problem

Consonants are every letter of the alphabet except a, e, i, o, and u.

Given a string, count how many consonants it contains.

Input

Programming

Output

Consonants: 8

Java Program

Java

public class CountConsonants {
    public static void main(String[] args) {
        String str = "Programming";
        String vowelSet = "aeiouAEIOU";
        int count = 0;
        for (char c : str.toCharArray()) {
            // A letter that isn't a vowel is a consonant
            if (Character.isLetter(c) && vowelSet.indexOf(c) == -1) count++;
        }
        System.out.println("Consonants: " + count);
    }
}

▶ Run Code

Output

Consonants: 8

Core Logic

A single pass through the string, keeping every letter that isn't a vowel, is the mirror image of counting vowels.

How It Works

1str.toCharArray() converts the string into a char[] so it can be scanned one character at a time.
2Character.isLetter(c) filters out spaces and punctuation, so only actual letters are considered.
3vowelSet.indexOf(c) == -1 checks that the letter is NOT one of "aeiouAEIOU".
4A letter passing both checks increments the count variable.

For "Programming", the scan counts P, r, g, r, m, m, n, g as consonants — eight in total.

💡

Key Point: Character.isLetter(c) has to run first — without it, digits and punctuation would also pass the 'not a vowel' check and get miscounted as consonants.

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()Character.isLetter()for-each loop

Approach 2: Regex Replace

Java

public class CountConsonantsRegex {
    public static void main(String[] args) {
        String str = "Programming";
        // Removes vowels and non-letters, leaving only consonants behind
        String consonantsOnly = str.replaceAll("(?i)[^bcdfghjklmnpqrstvwxyz]", "");
        System.out.println("Consonants: " + consonantsOnly.length());
    }
}

▶ Run Code

Output

Consonants: 8

Core Logic

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

How It Works

1str.replaceAll("(?i)[^bcdfghjklmnpqrstvwxyz]", "") matches anything that is NOT a consonant letter and removes it.
2(?i) makes the character class case-insensitive, so both uppercase and lowercase consonants are kept.
3What's left is a new string containing only the consonants from the original.
4.length() on that filtered string gives the total consonant count directly.

Replacing every non-consonant in "Programming" leaves "Prgrmmng", whose length is 8.

💡

Key Point: The character class explicitly lists every consonant letter rather than excluding vowels, since regex character classes match single characters against a fixed set, not against a 'not one of these five' rule directly stated as vowels.

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 CountConsonantsStream {
    public static void main(String[] args) {
        String str = "Programming";
        String vowelSet = "aeiouAEIOU";
        // Keeps only letters, then only the letters that aren't vowels
        long count = str.chars()
            .filter(Character::isLetter)
            .filter(c -> vowelSet.indexOf(c) == -1)
            .count();
        System.out.println("Consonants: " + count);
    }
}

▶ Run Code

Output

Consonants: 8

Core Logic

The same two-step filter — letters, then not-a-vowel — reads naturally as a chained stream pipeline.

How It Works

1str.chars() returns an IntStream of the string's character codes.
2.filter(Character::isLetter) keeps only letters, dropping spaces and punctuation.
3.filter(c -> vowelSet.indexOf(c) == -1) keeps only the letters that aren't vowels.
4.count() reduces the twice-filtered stream down to a single long total.

Filtering "Programming" down to letters, then to non-vowels, leaves eight characters — the same consonants the loop version found.

💡

Key Point: Chaining two .filter() calls reads as 'letters, then non-vowels' in the same order the manual version's && condition checks them.

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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