Is Valid Word easy or hard?

Valid Word is categorized as an Easy problem. It focuses on straightforward string iteration and condition checks rather than advanced algorithms, making it a common warm-up for string validation questions.

Valid Word Python/Java solution

Python and Java implementations both follow the same pattern: loop through the characters, check alphanumeric validity, and update vowel or consonant flags. After the scan, confirm that the string length is at least three and that both flags are true.

How to solve Valid Word in O(n)?

Iterate through the string once and classify each character. Reject the word if any character is not a digit or English letter. Maintain two flags for vowel and consonant detection; after the loop, return true only if both are present and the length is at least three.

What is the best approach for Valid Word?

The best approach is a single-pass iterative scan of the string. Track whether at least one vowel and one consonant appear while ensuring every character is alphanumeric. This runs in O(n) time and O(1) space and cleanly handles all constraints in one traversal.

Is Valid Word asked at Google/Amazon/Meta?

Problems like Valid Word appear in screening rounds at companies such as Amazon and Google because they test careful input validation and string processing. The logic is simple but requires handling edge cases such as digits, uppercase letters, and invalid characters.

What data structure is used in Valid Word?

No complex data structure is required. The solution relies on basic string traversal with constant variables to track vowel and consonant presence. Some implementations use a small set or lookup string for vowels to simplify checks.

What is the time complexity of Valid Word?

The time complexity is O(n), where n is the length of the input string. Each character is inspected exactly once to verify whether it is a letter or digit and to track vowel and consonant presence. Space complexity stays O(1) because only a few boolean flags are used.

Description Hints 1 Solution Video 9:00

3136. Valid Word

Easy50.9% acceptance8 companies

A word is considered valid if:

It contains a minimum of 3 characters.
It contains only digits (0-9), and English letters (uppercase and lowercase).
It includes at least one vowel.
It includes at least one consonant.

You are given a string word.

Return true if word is valid, otherwise, return false.

Notes:

'a', 'e', 'i', 'o', 'u', and their uppercases are vowels.
A consonant is an English letter that is not a vowel.

Example 1:

Input: word = "234Adas"

Output: true

Explanation:

This word satisfies the conditions.

Example 2:

Input: word = "b3"

Output: false

Explanation:

The length of this word is fewer than 3, and does not have a vowel.

Example 3:

Input: word = "a3$e"

Output: false

Explanation:

This word contains a '$' character and does not have a consonant.

Constraints:

1 <= word.length <= 20
word consists of English uppercase and lowercase letters, digits, '@', '#', and '$'.

Asked at 8 companies

Expedia UKG Microsoft Bloomberg

TCS

Google Amazon Meta

Topics

String

Hints open one at a time

Hint 1

Use if-else to check all the conditions.

9:00

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
2    def isValid(self, word: str) -> bool:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

"234Adas"

Input

"b3"

Input

"a3$e"

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