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Valid Word

Difficulty: Easy

Problem Description

A word is considered valid if it contains a minimum of 3 characters, contains only digits (0-9) and English letters (uppercase and lowercase), includes at least one vowel, and includes at least one consonant. You are given a string word. Return true if word is valid, otherwise, return false.

Key Insights

  • The word must have a length of at least 3 characters.
  • Valid characters include only digits and English letters.
  • At least one vowel (a, e, i, o, u) must be present in the word.
  • At least one consonant (any letter that is not a vowel) must be present in the word.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the input string. Space Complexity: O(1), as we are using a constant amount of space to store flags and counters.

Solution

To determine if a word is valid, we will:

  1. Check the length of the word to ensure it's at least 3 characters.
  2. Iterate through each character in the word to:
    • Check if the character is a valid letter or digit.
    • Track the presence of at least one vowel and one consonant.
  3. Return true if all conditions are met; otherwise, return false.

The algorithm primarily uses a single pass through the string while maintaining boolean flags for vowels and consonants.

Code Solutions

def is_valid_word(word: str) -> bool:
    if len(word) < 3:
        return False
    
    has_vowel = False
    has_consonant = False
    
    for char in word:
        if char.isalpha():  # Check if character is a letter
            if char.lower() in 'aeiou':
                has_vowel = True
            else:
                has_consonant = True
        elif not char.isdigit():  # Check if character is not a digit
            return False
    
    return has_vowel and has_consonant
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