EasyNo. 1119LeetCode Premium
Remove Vowels from a String — LeetCode 1119 Python Solution
- Pattern
- Hash Map
- Topics
- String
- Reading time
- 2 min
- Source
- leetcode.com
The problem
Given a string s, remove the vowels 'a', 'e', 'i', 'o', and 'u' from it, and return the new string.
Example
- Input
- s = "leetcodeisacommunityforcoders"
- Output
- "ltcdscmmntyfrcdrs"
Python solution
class Solution:
def removeVowels(self, s: str) -> str:
return "".join(c for c in s if c not in "aeiou")Complexity
- Time
- O(n)where n is the length of the string
- Space · auxiliary
- O(1)
Pattern: Hash Map
Trade memory for time: remember what you have seen so the second pass never happens. LeetCode 1119. Remove Vowels from a String is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.
The hash map guide has the Python template for the pattern and the 709 LeetCode problems that use it.
Related problems
| No. | Problem | Difficulty |
|---|---|---|
| 0006 | Zigzag Conversion | Medium |
| 0008 | String to Integer (atoi) | Medium |
| 0038 | Count and Say | Medium |
| 0058 | Length of Last Word | Easy |
| 0065 | Valid Number | Hard |
| 0068 | Text Justification | Hard |
Frequently asked questions
- How hard is LeetCode 1119. Remove Vowels from a String?
LeetCode 1119. Remove Vowels from a String is rated Easy on LeetCode.
- What is the time complexity of LeetCode 1119. Remove Vowels from a String?
The Python solution on this page runs in O(n), where n is the length of the string.
- What is the space complexity of LeetCode 1119. Remove Vowels from a String?
The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 1119. Remove Vowels from a String cover?
LeetCode 1119. Remove Vowels from a String is tagged String on LeetCode.
- Is LeetCode 1119. Remove Vowels from a String a premium problem?
Yes. LeetCode 1119. Remove Vowels from a String is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.