Java Programs›Strings›Remove Vowels from a String

Remove Vowels from a String in Java

beginner· Strings · String Manipulation

Problem

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

Given a string, remove every vowel it contains.

Input

Programming

Output

Prgrmmng

Java Program

Java

public class RemoveVowels {
    public static void main(String[] args) {
        String str = "Programming";
        String vowelSet = "aeiouAEIOU";
        StringBuilder result = new StringBuilder();
        for (char c : str.toCharArray()) {
            if (vowelSet.indexOf(c) == -1) result.append(c); // skip vowels entirely
        }
        System.out.println(result.toString());
    }
}

▶ Run Code

Output

Prgrmmng

Core Logic

Keeping every character that isn't in a small set of vowels, one pass through the string, filters out the vowels and leaves every consonant in place.

How It Works

1A for-each loop visits each character of the string in turn.
2vowelSet.indexOf(c) checks whether the current character appears in "aeiouAEIOU".
3Characters that don't appear in the vowel set are appended to a StringBuilder; vowels are simply skipped.
4The final StringBuilder holds every original character except the vowels.

For "Programming", the vowels 'o', 'a', and 'i' are skipped, producing "Prgrmmng".

💡

Key Point: This mirrors the vowel-counting program exactly, just appending non-vowels to a result instead of incrementing a counter for vowels.

Complexity

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

Why: Each character is visited once and the result buffer grows to hold every non-vowel character, which can be up to n.

Key Concepts

String.indexOf()StringBuilderfor-each loop

Approach 2: Regex Replace

Java

public class RemoveVowelsRegex {
    public static void main(String[] args) {
        String str = "Programming";
        // Matches and removes every vowel character, in both cases
        String result = str.replaceAll("[aeiouAEIOU]", "");
        System.out.println(result);
    }
}

▶ Run Code

Output

Prgrmmng

Core Logic

A regex character class listing all five vowels, in both cases, removes every match in one call.

How It Works

1str.replaceAll("[aeiouAEIOU]", "") matches any character that is one of the ten listed vowel characters.
2Every matched vowel is replaced with an empty string, which deletes it.
3What's left is a new string containing only the consonants and non-letters from the original.

Replacing every vowel in "Programming" removes 'o', 'a', and 'i', producing "Prgrmmng".

💡

Key Point: Unlike the negated character class used to strip special characters, this one lists the vowels directly, since there are only ten to match against.

Complexity

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

Why: replaceAll() still has to scan the whole string and build a new one holding every kept character.

Key Concepts

regexreplaceAll()

Approach 3: Java 8

Java

import java.util.stream.Collectors;

public class RemoveVowelsStream {
    public static void main(String[] args) {
        String str = "Programming";
        String vowelSet = "aeiouAEIOU";
        // Keeps every character that isn't a vowel, then joins them back into a String
        String result = str.chars()
            .filter(c -> vowelSet.indexOf(c) == -1)
            .mapToObj(c -> String.valueOf((char) c))
            .collect(Collectors.joining());
        System.out.println(result);
    }
}

▶ Run Code

Output

Prgrmmng

Core Logic

The same is-not-a-vowel check can filter a stream of character codes, then join what's left back into a string.

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 don't appear in the vowel set.
3.mapToObj(c -> String.valueOf((char) c)) converts each surviving code back into a one-character String.
4.collect(Collectors.joining()) concatenates them all back into a single result string.

Filtering "Programming" drops every vowel, and joining what's left reassembles "Prgrmmng".

💡

Key Point: This is the mirror image of the vowel-counting stream pipeline — same filter condition, but joining the survivors into a string instead of counting them.

Complexity

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

Why: The stream still visits every character once, and Collectors.joining() builds a result string holding every non-vowel character.

Key Concepts

Streamchars()filter()Collectors.joining()

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