Remove Vowels from String

This question is asked by Amazon. Given a string s, remove all the vowels it contains and return the resulting string.

Note: In this problem y is not considered a vowel.

Example(s)

Example 1:

Input: s = "aeiou"
Output: ""
Explanation:
All characters are vowels (a, e, i, o, u), so removing them results in an empty string.

Example 2:

Input: s = "byte"
Output: "byt"
Explanation:
- 'b' is not a vowel → keep
- 'y' is not a vowel (per problem note) → keep
- 't' is not a vowel → keep
- 'e' is a vowel → remove
Result: "byt"

Example 3:

Input: s = "xyz"
Output: "xyz"
Explanation:
- 'x' is not a vowel → keep
- 'y' is not a vowel → keep
- 'z' is not a vowel → keep
All characters are kept: "xyz"

Solution

The solution uses string filtering:

1. Define vowels: Create a set of vowels (a, e, i, o, u)
2. Filter characters: Keep only non-vowel characters
3. Return result: Return the filtered string

JavaScript Solution

JavaScript Solution - Filter

/**
* Remove all vowels from string
* @param {string} s - Input string
* @return {string} - String with vowels removed
*/
function removeVowels(s) {
// Define vowels (y is not considered a vowel)
const vowels = new Set(['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']);

// Filter out vowels
return s.split('').filter(char => !vowels.has(char)).join('');
}

// Test cases
console.log('Example 1:', removeVowels("aeiou")); 
// ""

console.log('Example 2:', removeVowels("byte")); 
// "byt"

console.log('Example 3:', removeVowels("xyz")); 
// "xyz"

console.log('Test 4:', removeVowels("leetcode")); 
// "ltcd"

Output:

Click "Run Code" to execute the code and see the results.

Python Solution

Python Solution - Filter

def remove_vowels(s: str) -> str:
  """
  Remove all vowels from string
  
  Args:
      s: Input string
      
  Returns:
      str: String with vowels removed
  """
  # Define vowels (y is not considered a vowel)
  vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
  
  # Filter out vowels
  return ''.join(char for char in s if char not in vowels)

# Test cases
print('Example 1:', remove_vowels("aeiou"))  
# ""

print('Example 2:', remove_vowels("byte"))  
# "byt"

print('Example 3:', remove_vowels("xyz"))  
# "xyz"

print('Test 4:', remove_vowels("leetcode"))  
# "ltcd"

Loading Python runtime...

Output:

Click "Run Code" to execute the code and see the results.

Alternative Solution (String Replacement)

Here's a version using string replacement:

JavaScript Replace

/**
 * Using string replacement
 */
function removeVowelsReplace(s) {
  return s.replace(/[aeiouAEIOU]/g, '');
}

Python Replace

import re

def remove_vowels_replace(s: str) -> str:
    """
    Using string replacement
    """
    return re.sub(r'[aeiouAEIOU]', '', s)

Alternative Solution (String Builder)

Here's a version using string building:

JavaScript Builder

/**
 * Using string builder
 */
function removeVowelsBuilder(s) {
  const vowels = new Set(['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']);
  let result = '';
  
  for (const char of s) {
    if (!vowels.has(char)) {
      result += char;
    }
  }
  
  return result;
}

Python Builder

def remove_vowels_builder(s: str) -> str:
    """
    Using string builder
    """
    vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
    result = []
    
    for char in s:
        if char not in vowels:
            result.append(char)
    
    return ''.join(result)

Complexity

• Time Complexity: O(n) - Where n is the length of the string. We process each character once.
• Space Complexity: O(n) - For the result string (or O(1) if modifying in-place, but strings are immutable in most languages).

Approach

The solution uses string filtering:

1. Define vowels:

   • Create a set of vowels: a, e, i, o, u (both lowercase and uppercase)
   • Note: y is NOT considered a vowel

2. Filter characters:

   • Iterate through each character
   • Keep only characters that are not vowels

3. Return result:

   • Join the filtered characters into a string
   • Return the result

Key Insights

• Set for vowels: Efficient O(1) lookup for vowel checking
• Filter approach: Clean and readable
• Case insensitive: Handle both uppercase and lowercase vowels
• Y is not a vowel: Important constraint in this problem
• O(n) time: Must process each character

Step-by-Step Example

Let's trace through Example 2: s = "byte"

s = "byte"
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}

Process each character:
  'b': vowels.has('b')? No → keep
  'y': vowels.has('y')? No → keep (y is not a vowel)
  't': vowels.has('t')? No → keep
  'e': vowels.has('e')? Yes → remove

Filtered: ['b', 'y', 't']
Join: "byt"

Result: "byt"

Visual Representation

Example 1: s = "aeiou"

Characters: a  e  i  o  u
Vowels?    ✓  ✓  ✓  ✓  ✓
Keep?      ✗  ✗  ✗  ✗  ✗

Result: "" (all removed)

Example 2: s = "byte"

Characters: b  y  t  e
Vowels?    ✗  ✗  ✗  ✓
Keep?      ✓  ✓  ✓  ✗

Result: "byt"

Example 3: s = "xyz"

Characters: x  y  z
Vowels?    ✗  ✗  ✗
Keep?      ✓  ✓  ✓

Result: "xyz" (all kept)

Edge Cases

• All vowels: Returns empty string
• No vowels: Returns original string
• Empty string: Returns empty string
• Mixed case: Handles both uppercase and lowercase
• Single character: Works correctly
• Only y: Returns "y" (y is not a vowel)

Important Notes

• Y is not a vowel: Important constraint in this problem
• Case insensitive: Handle both uppercase and lowercase
• Set lookup: O(1) time for vowel checking
• O(n) time: Must process each character
• O(n) space: For the result string

Vowel Definition

Standard vowels: a, e, i, o, u

• Both lowercase and uppercase: A, E, I, O, U
• Y is NOT a vowel in this problem (unlike some contexts)

Related Problems

• Reverse Vowels of a String: Different problem
• Longest Substring Without Repeating Characters: Different problem
• Valid Anagram: Different string problem
• Remove Duplicates: Different problem

Takeaways

• Set for vowels provides efficient O(1) lookup
• Filter approach is clean and readable
• Y is not a vowel is an important constraint
• O(n) time is optimal (must process each character)
• Case handling for both uppercase and lowercase