LeetCode 3136: Valid Word

Meta Data

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Difficulty: easy First Attempt: 2024-05-05

• Submissions: Wrong Answer → Accepted → Wrong Answer → Accepted

Approach

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Check the four conditions one by one (String Template 4): length >= 3, every character alphanumeric, at least one vowel (either case), and at least one consonant. A consonant is a letter that is not a vowel; digits do not count.

My Attempts

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• The first 2024 version treated every non-vowel character as a consonant, so digits counted: "a3E" has only vowels and a digit, so the answer is False; it returned True.
• A 2025 rewrite tested vowels with w in "aeiou", missing uppercase vowels: "bEE" returned False instead of True. Using w.lower() passed.

Solution

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class Solution:
    def isValid(self, word: str) -> bool:
        if len(word) < 3 or not all(c.isalnum() for c in word):
            return False
        vowels = set("aeiouAEIOU")
        has_vowel = any(c in vowels for c in word)
        has_consonant = any(c.isalpha() and c not in vowels for c in word)
        return has_vowel and has_consonant

Complexity

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Time O(n), space O(1).

Mistakes to Avoid

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• Digits are not consonants; check that the character is a letter.
• Uppercase vowels are vowels too.

Interview Explanation

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I check length at least 3 and all characters alphanumeric, then require a vowel and a consonant, where a consonant must be a non-vowel letter. Time O(n).