Trim Trailing Vowels

Difficulty: Easy | Acceptance: 77.40% | Paid: No Topics: String

Given a string s, return the string after removing all trailing vowels. A trailing vowel is a vowel that appears at the end of the string. Vowels are ‘a’, ‘e’, ‘i’, ‘o’, ‘u’ (both lowercase and uppercase).

• Examples
• Constraints
• Two Pointer Approach
• Built-in Methods Approach
• Iterative Approach

Examples

Example 1

Input:

s = "idea"

Output:

"id"

Explanation: Removing “idea”, we obtain the string “id”.

Example 2

Input:

s = "day"

Output:

"day"

Explanation: There are no trailing vowels in the string “day”.

Example 3

Input:

s = "aeiou"

Output:

""

Explanation: Removing “aeiou”, we obtain the string "".

Constraints

1 <= s.length <= 10⁵
s consists of lowercase and uppercase English letters.

Two Pointer Approach

Intuition Start from the end of the string and move backwards until we find a non-vowel character, then return the substring up to that point.

Steps

• Initialize a pointer at the end of the string
• Move the pointer backwards while the current character is a vowel
• Return the substring from the start to the pointer position

class Solution:
    def trimTrailingVowels(self, s: str) -> str:
        vowels = set('aeiouAEIOU')
        i = len(s) - 1
        while i >= 0 and s[i] in vowels:
            i -= 1
        return s[:i+1]

Complexity

• Time: O(n) where n is the length of the string
• Space: O(1) extra space (not counting the output string)
• Notes: Simple and efficient approach with minimal space usage

Built-in Methods Approach

Intuition Use language-specific built-in methods to remove trailing characters that match a pattern.

Steps

• Use regex or built-in string methods to remove trailing vowels
• Return the modified string

import re

class Solution:
    def trimTrailingVowels(self, s: str) -> str:
        return re.sub(r'[aeiouAEIOU]+$', '', s)

Complexity

• Time: O(n) where n is the length of the string
• Space: O(n) for the new string (in most languages)
• Notes: Concise code using regex, but may have slightly higher overhead

Iterative Approach

Intuition Build the result string by iterating from the end and collecting characters until we hit a non-vowel.

Steps

• Start from the end of the string
• Collect characters until we find a non-vowel
• Return the result substring

class Solution:
    def trimTrailingVowels(self, s: str) -> str:
        vowels = set('aeiouAEIOU')
        i = len(s) - 1
        while i >= 0 and s[i] in vowels:
            i -= 1
        return s[:i+1]

Complexity

• Time: O(n) where n is the length of the string
• Space: O(1) extra space (not counting the output string)
• Notes: Similar to two-pointer approach but with explicit iteration