Valid Word

Problem

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.

Examples

Example 1

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Input: word = "234Adas"
Output: true
Explanation:
This word satisfies the conditions.

Example 2

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Input: word = "b3"
Output: false
Explanation:
The length of this word is fewer than 3, and does not have a vowel.

Example 3

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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 '$'.

Solution

Method 1 – Simple Checks

Intuition

We need to check the length, allowed characters, and presence of at least one vowel and one consonant.

Approach

Check if the length is at least 3.
Check if all characters are alphanumeric (letters or digits).
Check for at least one vowel and at least one consonant.

Code

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bool isValid(string word) {
    if (word.length() < 3) return false;
    string vowels = "aeiouAEIOU";
    bool hasVowel = false, hasConsonant = false;
    for (char c : word) {
        if (!isalnum(c)) return false;
        if (isalpha(c)) {
            if (vowels.find(c) != string::npos) hasVowel = true;
            else hasConsonant = true;
        }
    }
    return hasVowel && hasConsonant;
}

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import "unicode"
func isValid(word string) bool {
    if len(word) < 3 {
        return false
    }
    vowels := "aeiouAEIOU"
    hasVowel, hasConsonant := false, false
    for _, c := range word {
        if !unicode.IsLetter(c) && !unicode.IsDigit(c) {
            return false
        }
        if unicode.IsLetter(c) {
            if strings.ContainsRune(vowels, c) {
                hasVowel = true
            } else {
                hasConsonant = true
            }
        }
    }
    return hasVowel && hasConsonant
}

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class Solution {
	public boolean isValid(String word) {
	    if (word.length() < 3) return false;
	    String vowels = "aeiouAEIOU";
	    boolean hasVowel = false, hasConsonant = false;
	    for (char c : word.toCharArray()) {
	        if (!Character.isLetterOrDigit(c)) return false;
	        if (Character.isLetter(c)) {
	            if (vowels.indexOf(c) >= 0) hasVowel = true;
	            else hasConsonant = true;
	        }
	    }
	    return hasVowel && hasConsonant;
	}
}

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fun isValid(word: String): Boolean {
    if (word.length < 3) return false
    val vowels = "aeiouAEIOU"
    var hasVowel = false
    var hasConsonant = false
    for (c in word) {
        if (!c.isLetterOrDigit()) return false
        if (c.isLetter()) {
            if (vowels.contains(c)) hasVowel = true
            else hasConsonant = true
        }
    }
    return hasVowel && hasConsonant
}

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def isValid(word: str) -> bool:
    if len(word) < 3:
        return False
    vowels = set('aeiouAEIOU')
    has_vowel = False
    has_consonant = False
    for c in word:
        if not c.isalnum():
            return False
        if c.isalpha():
            if c in vowels:
                has_vowel = True
            else:
                has_consonant = True
    return has_vowel and has_consonant

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pub fn is_valid(word: &str) -> bool {
    if word.len() < 3 { return false; }
    let vowels = "aeiouAEIOU";
    let (mut has_vowel, mut has_consonant) = (false, false);
    for c in word.chars() {
        if !c.is_ascii_alphanumeric() { return false; }
        if c.is_ascii_alphabetic() {
            if vowels.contains(c) { has_vowel = true; }
            else { has_consonant = true; }
        }
    }
    has_vowel && has_consonant
}

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function isValid(word: string): boolean {
    if (word.length < 3) return false;
    const vowels = "aeiouAEIOU";
    let hasVowel = false, hasConsonant = false;
    for (const c of word) {
        if (!/[a-zA-Z0-9]/.test(c)) return false;
        if (/[a-zA-Z]/.test(c)) {
            if (vowels.includes(c)) hasVowel = true;
            else hasConsonant = true;
        }
    }
    return hasVowel && hasConsonant;
}

Complexity

⏰ Time complexity: O(n) — n is the length of the word.
🧺 Space complexity: O(1) — Only a few variables are used.

Problem#

Examples#

Example 1#

Example 2#

Example 3#

Constraints#

Solution#

Method 1 – Simple Checks#

Intuition#

Approach#

Code#

Complexity#