Valid Word

easy string validation vowels

Problem

A word is valid when it satisfies all of these rules: it has a minimum of 3 characters; it contains only digits (0–9) and English letters (upper or lower case); it includes at least one vowel (a, e, i, o, u and their uppercases); and it includes at least one consonant (any English letter that is not a vowel). Given a string word, return true if it is valid, otherwise false.

Inputword = "234Adas"

Outputtrue

Length 7 ≥ 3, every character is a digit or letter, ‘A’ and ‘a’ are vowels, and ‘d’, ‘s’ are consonants — all conditions hold.

word

def isValid(word):
    if len(word) < 3:                       # rule 1: at least 3 characters
        return False
    vowels = set("aeiouAEIOU")
    has_vowel = False
    has_consonant = False
    for ch in word:                         # scan every character
        if ch.isalpha():                    # an English letter
            if ch in vowels:
                has_vowel = True
            else:
                has_consonant = True
        elif not ch.isdigit():              # not a letter, not a digit
            return False                    # illegal character (e.g. '$')
    return has_vowel and has_consonant      # need a vowel AND a consonant

function isValid(word) {
  if (word.length < 3) return false;        // rule 1: at least 3 characters
  const vowels = new Set([..."aeiouAEIOU"]);
  let hasVowel = false, hasConsonant = false;
  for (const ch of word) {                   // scan every character
    if (/[a-zA-Z]/.test(ch)) {               // an English letter
      if (vowels.has(ch)) hasVowel = true;
      else hasConsonant = true;
    } else if (!/[0-9]/.test(ch)) {          // not a letter, not a digit
      return false;                          // illegal character (e.g. '$')
    }
  }
  return hasVowel && hasConsonant;          // need a vowel AND a consonant
}

boolean isValid(String word) {
    if (word.length() < 3) return false;     // rule 1: at least 3 characters
    String vowels = "aeiouAEIOU";
    boolean hasVowel = false, hasConsonant = false;
    for (char ch : word.toCharArray()) {     // scan every character
        if (Character.isLetter(ch)) {        // an English letter
            if (vowels.indexOf(ch) >= 0) hasVowel = true;
            else hasConsonant = true;
        } else if (!Character.isDigit(ch)) { // not a letter, not a digit
            return false;                    // illegal character (e.g. '$')
        }
    }
    return hasVowel && hasConsonant;         // need a vowel AND a consonant
}

bool isValid(string word) {
    if (word.size() < 3) return false;       // rule 1: at least 3 characters
    string vowels = "aeiouAEIOU";
    bool hasVowel = false, hasConsonant = false;
    for (char ch : word) {                   // scan every character
        if (isalpha((unsigned char)ch)) {    // an English letter
            if (vowels.find(ch) != string::npos) hasVowel = true;
            else hasConsonant = true;
        } else if (!isdigit((unsigned char)ch)) { // not a letter, not a digit
            return false;                    // illegal character (e.g. '$')
        }
    }
    return hasVowel && hasConsonant;         // need a vowel AND a consonant
}

Explanation

This is a pure validation problem: there is no clever data structure, just a careful check of four independent rules over a short string (length up to 20).

We start with the cheapest test — the length. If the word has fewer than 3 characters it can never be valid, so we return false immediately and avoid scanning anything.

Then we make a single pass over the characters, carrying two boolean flags: has_vowel and has_consonant. For each character we classify it. If it is an English letter, we ask whether it is one of aeiouAEIOU: a match sets the vowel flag, anything else sets the consonant flag. If it is a digit we simply ignore it — digits are allowed but contribute to neither flag.

If a character is neither a letter nor a digit (for example '$', '@', or '#'), the word breaks the “only digits and letters” rule, so we can stop early and return false.

After the scan, the word is valid exactly when both flags are true: at least one vowel and at least one consonant were seen. We return has_vowel and has_consonant.

Example: "a3$e" is length 4 and has vowels 'a', 'e', but the '$' is an illegal character — and there is no consonant either — so the answer is false.

Time: O(n) Space: O(1)