Vowel Counter - Problem

Given a string, count the number of vowels and consonants in it. Ignore any non-alphabetic characters (numbers, spaces, punctuation).

A vowel is one of: a, e, i, o, u (case-insensitive).

Return an array where the first element is the vowel count and the second element is the consonant count.

Input & Output

Example 1 — Basic Mixed String

$ Input: s = "Hello World!"

› Output: [2, 3]

💡 Note: Letters: H(consonant), e(vowel), l(consonant), l(consonant), o(vowel). Ignore space, exclamation mark, and 'W', 'o', 'r', 'l', 'd' from 'World' gives us more letters. Total: 2 vowels (e, o), 3 consonants (H, l, l) from 'Hello', plus 'W'(consonant), 'r'(consonant), 'l'(consonant), 'd'(consonant) from 'World'. Final count: 3 vowels (e, o, o), 7 consonants (H, l, l, W, r, l, d). Actually: 'Hello World!' has vowels: e, o, o = 3 vowels; consonants: H, l, l, W, r, l, d = 7 consonants. Wait, let me recount: H-e-l-l-o W-o-r-l-d = vowels(e,o,o) = 3, consonants(H,l,l,W,r,l,d) = 7. So output should be [3,7].

Example 2 — Only Vowels

$ Input: s = "aeiou"

› Output: [5, 0]

💡 Note: All characters are vowels: a, e, i, o, u. No consonants present.

Example 3 — Mixed with Numbers

$ Input: s = "abc123def"

› Output: [2, 4]

💡 Note: Letters only: a(vowel), b(consonant), c(consonant), d(consonant), e(vowel), f(consonant). Ignore numbers 123. Total: 2 vowels, 4 consonants.

Constraints

1 ≤ s.length ≤ 10⁴
s consists of printable ASCII characters

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use a set or hash table for O(1) vowel lookups instead of searching through a string each time. Best approach creates a vowel set and makes one pass through the input string. Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force Check

⏱️ Time: O(n) Space: O(1)

Iterate through each character in the string. For each character, first check if it's alphabetic. If yes, check if it's a vowel by comparing against 'aeiouAEIOU' string every time.

Set-Based Lookup

⏱️ Time: O(n) Space: O(1)

Create a set of vowels at the start for instant lookup. Iterate through the string once, checking each alphabetic character against the vowel set in constant time.

Brute Force Check — Algorithm Steps

Step 1: Initialize vowel and consonant counters to 0
Step 2: For each character, check if it's alphabetic
Step 3: If alphabetic, search through 'aeiouAEIOU' to check if it's a vowel
Step 4: Increment appropriate counter

Visualization

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Step-by-Step Walkthrough

Check Character

Examine if character is alphabetic

Search Vowels

Linear search through 'aeiouAEIOU' string

Count

Increment vowel or consonant counter

Code -

solution.c — C

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

int* solution(char* s) {
    int* result = malloc(2 * sizeof(int));
    result[0] = 0; // vowel count
    result[1] = 0; // consonant count
    
    char* vowels = "aeiouAEIOU";
    
    for (int i = 0; s[i] != '\0'; i++) {
        if (isalpha(s[i])) {
            if (strchr(vowels, s[i]) != NULL) {
                result[0]++;
            } else {
                result[1]++;
            }
        }
    }
    
    return result;
}

int main() {
    char s[10001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int* result = solution(s);
    printf("[%d,%d]\n", result[0], result[1]);
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through string, vowel check is constant time for 10 vowel characters

✓ Linear Growth

Space Complexity

O(1)

Only using two counter variables

✓ Linear Space

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Vowel Counter - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Check — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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