How do you solve Trim Trailing Vowels on LeetCode?

Trim Trailing Vowels (LeetCode #3856) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Trailing vowels can be efficiently trimmed by scanning from the end.

What is the time complexity of Trim Trailing Vowels?

The optimal solution for Trim Trailing Vowels runs in O(n) time and uses O(n) space. The complexity is O(n) because we traverse the string once, and O(n) space is used for the substring.

What is the brute force solution for Trim Trailing Vowels?

The brute force approach for Trim Trailing Vowels is Brute Force, with O(n²) time complexity and O(1) space complexity. We can remove trailing vowels by checking each character from the end of the string until we find a non-vowel. This simulates the process of trimming. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Trim Trailing Vowels?

Trim Trailing Vowels uses the following concepts: String. The recommended patterns to study are: String Manipulation, Two Pointers.

What are the interview tips for Trim Trailing Vowels?

Explain your thought process clearly. Consider edge cases during your solution. Optimize your solution after the brute force approach.

What are common mistakes when solving Trim Trailing Vowels?

Not considering edge cases like all vowels. Incorrectly indexing the string when returning the result.

#3856

Trim Trailing Vowels

Easy

StringString ManipulationTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can efficiently find the last non-vowel character by scanning the string from the end, making it a linear time solution.

⚙️

Algorithm

3 steps

1Step 1: Initialize a pointer at the end of the string.
2Step 2: Move the pointer left until a non-vowel is found.
3Step 3: Return the substring from the start to the pointer index.

solution.py5 lines

1def trim_trailing_vowels(s):
2    i = len(s) - 1
3    while i >= 0 and s[i] in 'aeiou':
4        i -= 1
5    return s[:i + 1]

ℹ

Complexity note: The complexity is O(n) because we traverse the string once, and O(n) space is used for the substring.

1Trailing vowels can be efficiently trimmed by scanning from the end.
2Understanding string manipulation is crucial for optimizing performance.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.