How do you solve Valid Word on LeetCode?

Valid Word (LeetCode #3136) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding character types (vowels, consonants, digits) is crucial for this problem.

What is the brute force solution for Valid Word?

The brute force approach for Valid Word is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach checks each condition for a valid word one by one. It iterates through the string to verify the length, character types, and presence of vowels and consonants. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Valid Word?

Valid Word uses the following concepts: String. The recommended patterns to study are: Character classification, Single-pass validation.

What are the interview tips for Valid Word?

Clearly explain your thought process while coding. Test your solution with edge cases, like strings with special characters. Remember to check the length condition first to avoid unnecessary checks.

What are common mistakes when solving Valid Word?

Not checking for non-alphanumeric characters properly. Failing to account for both vowel and consonant presence.

#3136

Valid Word

Easy

StringCharacter classificationSingle-pass validation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the string to check all conditions simultaneously. This reduces the number of iterations and checks, making it efficient.

⚙️

Algorithm

6 steps

1Step 1: Check if the length of the word is less than 3. If yes, return false.
2Step 2: Initialize flags for vowel and consonant presence.
3Step 3: Iterate through each character in the word and check its type in one go.
4Step 4: If a character is a letter, check if it's a vowel or consonant and update the flags.
5Step 5: If a character is not a digit or letter, return false immediately.
6Step 6: After the loop, return true if both flags are true; otherwise, return false.

solution.py13 lines

1def isValidWord(word):
2    if len(word) < 3:
3        return False
4    has_vowel = has_consonant = False
5    for char in word:
6        if char.isalpha():
7            if char.lower() in 'aeiou':
8                has_vowel = True
9            else:
10                has_consonant = True
11        elif not char.isdigit():
12            return False
13    return has_vowel and has_consonant

ℹ

Complexity note: The time complexity is O(n) as we traverse the string once. The space complexity is O(1) since we only use a couple of boolean flags.

1Understanding character types (vowels, consonants, digits) is crucial for this problem.
2Checking conditions simultaneously can improve efficiency.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.