Valid Word - Practice Coding Problems

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: true

💡 Note: Length is 4 (≥3), contains only alphanumeric characters, has vowels 'U', 'u', 'E' and no consonants, so returns false. Wait - let me recalculate: 'U', 'u', 'E' are vowels but there's no consonant, so this should be 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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Understanding the Visualization

Input

String word to validate

Validate

Check length, characters, vowels, consonants

Output

Boolean result

Key Takeaway

🎯 Key Insight: Single pass validation checking all four conditions simultaneously

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

Approach	Time	Space	Notes
✓ Single Pass Validation	O(n)	O(1)	Check all conditions in one pass through the string
Early Termination Optimization	O(n)	O(1)	Stop checking once all conditions are satisfied

Single Pass Validation — Algorithm Steps

Step 1: Check if length is at least 3
Step 2: For each character, verify it's alphanumeric
Step 3: Track if we've seen vowels and consonants
Step 4: Return true only if all conditions are met

Visualization

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Step-by-Step Walkthrough

Check Length

Verify word has at least 3 characters

Validate Characters

Each character must be alphanumeric

Track Vowels/Consonants

Count presence of vowels and consonants

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)

Single pass through string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

12.0K Views

Medium Frequency

~15 min Avg. Time

450 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass Validation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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