Leetcode #2785: Sort Vowels in a String
In this guide, we solve Leetcode #2785 Sort Vowels in a String in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a 0-indexed string s, permute s to get a new string t such that: All consonants remain in their original places. More formally, if there is an index i with 0 <= i < s.length such that s[i] is a consonant, then t[i] = s[i].
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: String, Sorting
Intuition
Sorting reveals structure that is hard to see in the original order.
Once sorted, a linear scan is usually enough to compute the answer.
Approach
Sort the data and sweep through it while maintaining a small state.
This keeps the logic straightforward and reliable.
Steps:
- Sort the data.
- Scan in order while maintaining state.
- Update the best answer.
Example
Input: s = "lEetcOde"
Output: "lEOtcede"
Explanation: 'E', 'O', and 'e' are the vowels in s; 'l', 't', 'c', and 'd' are all consonants. The vowels are sorted according to their ASCII values, and the consonants remain in the same places.
Python Solution
class Solution:
def sortVowels(self, s: str) -> str:
vs = [c for c in s if c.lower() in "aeiou"]
vs.sort()
cs = list(s)
j = 0
for i, c in enumerate(cs):
if c.lower() in "aeiou":
cs[i] = vs[j]
j += 1
return "".join(cs)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.