LeetCode 3775: Reverse Words With Same Vowel Count ·

Table of Contents

Meta Data
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Difficulty: medium First Attempt: 2026-01-24

Submissions: Accepted (first try)

Approach
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Split the string on spaces.
Count the vowels of the first word, target.
Leave the first word alone; reverse each later word whose vowel count equals target.

Reversing a word can use slicing, word[::-1], or converging pointers.

My Solution
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class Solution:
    def reverseWords(self, s: str) -> str:
        def vowels(word: str) -> int:
            return sum(ch in "aeiou" for ch in word)

        words = s.split(" ")
        target = vowels(words[0])
        return " ".join([words[0]] + [w[::-1] if vowels(w) == target else w for w in words[1:]])

My original counted vowels by calling word.count(v) once per vowel, scanning the word five times; this version scans once. It also detected the first word by checking whether res was empty; here the first word is handled separately.

Complexity
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Time O(n), space O(n).

Mistakes to Avoid
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Reversing the first word too: only the words after it are processed.

Interview Explanation
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I split on spaces, count the first word’s vowels, reverse each later word with the same count and keep the others, then join with spaces. Each character is looked at a constant number of times, so time is O(n).