LeetCode #3136 — EASY

Valid Word

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

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The Problem

Problem Statement

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 '$'.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"234Adas"

Example 2

"b3"

Example 3

"a3$e"

Step 02

Core Insight

What unlocks the optimal approach

Use if-else to check all the conditions.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3136: Valid Word
class Solution {
    public boolean isValid(String word) {
        if (word.length() < 3) {
            return false;
        }
        boolean hasVowel = false, hasConsonant = false;
        boolean[] vs = new boolean[26];
        for (char c : "aeiou".toCharArray()) {
            vs[c - 'a'] = true;
        }
        for (char c : word.toCharArray()) {
            if (Character.isAlphabetic(c)) {
                if (vs[Character.toLowerCase(c) - 'a']) {
                    hasVowel = true;
                } else {
                    hasConsonant = true;
                }
            } else if (!Character.isDigit(c)) {
                return false;
            }
        }
        return hasVowel && hasConsonant;
    }
}

// Accepted solution for LeetCode #3136: Valid Word
func isValid(word string) bool {
	if len(word) < 3 {
		return false
	}
	hasVowel := false
	hasConsonant := false
	vs := make([]bool, 26)
	for _, c := range "aeiou" {
		vs[c-'a'] = true
	}
	for _, c := range word {
		if unicode.IsLetter(c) {
			if vs[unicode.ToLower(c)-'a'] {
				hasVowel = true
			} else {
				hasConsonant = true
			}
		} else if !unicode.IsDigit(c) {
			return false
		}
	}
	return hasVowel && hasConsonant
}

# Accepted solution for LeetCode #3136: Valid Word
class Solution:
    def isValid(self, word: str) -> bool:
        if len(word) < 3:
            return False
        has_vowel = has_consonant = False
        vs = set("aeiouAEIOU")
        for c in word:
            if not c.isalnum():
                return False
            if c.isalpha():
                if c in vs:
                    has_vowel = True
                else:
                    has_consonant = True
        return has_vowel and has_consonant

// Accepted solution for LeetCode #3136: Valid Word
impl Solution {
    pub fn is_valid(word: String) -> bool {
        if word.len() < 3 {
            return false;
        }

        let mut has_vowel = false;
        let mut has_consonant = false;
        let vowels = ['a', 'e', 'i', 'o', 'u'];

        for c in word.chars() {
            if !c.is_alphanumeric() {
                return false;
            }
            if c.is_alphabetic() {
                let lower_c = c.to_ascii_lowercase();
                if vowels.contains(&lower_c) {
                    has_vowel = true;
                } else {
                    has_consonant = true;
                }
            }
        }

        has_vowel && has_consonant
    }
}

// Accepted solution for LeetCode #3136: Valid Word
function isValid(word: string): boolean {
    if (word.length < 3) {
        return false;
    }
    let hasVowel: boolean = false;
    let hasConsonant: boolean = false;
    const vowels: Set<string> = new Set(['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']);
    for (const c of word) {
        if (!c.match(/[a-zA-Z0-9]/)) {
            return false;
        }
        if (/[a-zA-Z]/.test(c)) {
            if (vowels.has(c)) {
                hasVowel = true;
            } else {
                hasConsonant = true;
            }
        }
    }
    return hasVowel && hasConsonant;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.