Problem

Given a string s, remove the vowels 'a', 'e', 'i', 'o', and 'u' from it, and return the new string.

Examples

Example 1:

1
2

Input: s = "leetcodeisacommunityforcoders"
Output: "ltcdscmmntyfrcdrs"

Example 2:

1
2

Input: s = "aeiou"
Output: ""

Constraints:

  • 1 <= s.length <= 1000
  • s consists of only lowercase English letters.

Solution

Method 1 - String Filtering

Intuition

We want to remove all vowels from the string. This can be done by iterating through the string and skipping any character that is a vowel.

Approach

  1. Define a set of vowels.
  2. Build a new string by including only non-vowel characters.

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

#include <string>
using namespace std;

string removeVowels(string s) {
    string res;
    for (char c : s) {
        if (c != 'a' && c != 'e' && c != 'i' && c != 'o' && c != 'u') res += c;
    }
    return res;
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10

func removeVowels(s string) string {
    vowels := map[byte]bool{'a': true, 'e': true, 'i': true, 'o': true, 'u': true}
    res := make([]byte, 0, len(s))
    for i := 0; i < len(s); i++ {
        if !vowels[s[i]] {
            res = append(res, s[i])
        }
    }
    return string(res)
}
1
2
3
4
5
6
7
8
9

public class Solution {
    public String removeVowels(String s) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (c != 'a' && c != 'e' && c != 'i' && c != 'o' && c != 'u') sb.append(c);
        }
        return sb.toString();
    }
}
1
2
3

fun removeVowels(s: String): String {
    return s.filter { it !in "aeiou" }
}
1
2

def removeVowels(s: str) -> str:
    return ''.join(c for c in s if c not in 'aeiou')
1
2
3

fn remove_vowels(s: &str) -> String {
    s.chars().filter(|&c| !"aeiou".contains(c)).collect()
}
1
2
3

function removeVowels(s: string): string {
    return s.split('').filter(c => !'aeiou'.includes(c)).join('');
}

Complexity

  • ⏰ Time complexity: O(N), where N is the length of s.
  • 🧺 Space complexity: O(N), for the output string.