Valid Word

Valid Word - Problem

String Easy

A word is considered valid if it satisfies the following conditions:

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.

Note: 'a', 'e', 'i', 'o', 'u' and their uppercases are vowels. A consonant is an English letter that is not a vowel.

Input & Output

Example 1 — Valid Word

$ Input: word = "UuE6"

› Output: false

💡 Note: Length is 4 (≥3), contains only alphanumeric characters, has vowels 'U', 'u', 'E' but no consonants, so returns false.

Example 2 — Invalid Word (No Consonant)

$ Input: word = "aeiou"

› Output: false

💡 Note: Has vowels but no consonants, so it's invalid.

Example 3 — Valid Mixed Word

$ Input: word = "a1B"

› Output: true

💡 Note: Length is 3, has vowel 'a', consonant 'B', and digit '1' - all conditions satisfied.

Constraints

1 ≤ word.length ≤ 20
word consists of English letters, digits, and special characters

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is to validate all conditions in a single pass: check length first, then for each character verify it's alphanumeric while tracking vowels and consonants. Best approach uses single pass validation with O(n) time and O(1) space.

Common Approaches

✓ Early Termination Optimization

⏱️ Time: O(n) Space: O(1)

Same validation logic but exit early when we've found at least one vowel and one consonant, potentially saving iterations.

Single Pass Validation

⏱️ Time: O(n) Space: O(1)

Iterate through each character once, validating length, character types, and counting vowels and consonants simultaneously.

Early Termination Optimization — Algorithm Steps

Step 1: Check length requirement first
Step 2: Track vowel and consonant flags
Step 3: Exit early if both flags are true and character is valid
Step 4: Continue validation until end or early exit

Visualization

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Step-by-Step Walkthrough

Check Characters

Validate and categorize each character

Track Progress

Monitor if vowel and consonant found

Continue Validation

Still must validate remaining characters

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdbool.h>

bool isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ||
           c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U';
}

bool solution(char* word) {
    int len = strlen(word);
    if (len < 3) {
        return false;
    }
    
    bool hasVowel = false;
    bool hasConsonant = false;
    
    for (int i = 0; i < len; i++) {
        if (!isalnum(word[i])) {
            return false;
        }
        
        if (isalpha(word[i])) {
            if (isVowel(word[i])) {
                hasVowel = true;
            } else {
                hasConsonant = true;
            }
        }
    }
    
    return hasVowel && hasConsonant;
}

int main() {
    char word[1001];
    fgets(word, sizeof(word), stdin);
    word[strcspn(word, "\n")] = 0;
    
    bool result = solution(word);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Still need to check all characters in worst case

✓ Linear Growth

Space Complexity

O(1)

Only constant extra space for flags

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Early Termination Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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