Count Vowels – Solution & Complexity

Q: What is the time and space complexity of Count Vowels?

The optimal Count Vowels solution runs in `O(n)` time and `O(1)` space. Scan the string once and increment a counter on each vowel.

Q: What is the most efficient way to solve Count Vowels?

Scan the string once and increment a counter on each vowel. This gives `O(n)` time complexity, and is a classic strings problem.

Q: How do you solve Count Vowels in Python & JavaScript?

We provide a worked Count Vowels solution in Python & JavaScript with a step-by-step explanation above. You can also open the interactive editor to write and run your own Python solution against the test cases in the browser.

Q: Is Count Vowels an easy or hard interview question?

Count Vowels is generally rated **Easy**. It tests strings, which are common coding-interview patterns.

1. Understand What Counts

Count the vowels a, e, i, o, and u.
The check is case-insensitive, so uppercase vowels count too.
Consonants, spaces, digits, and punctuation do not add to the total.

2. Normalize Characters

To avoid checking both uppercase and lowercase forms, convert each character to lowercase.
Then compare it with a small set of vowel characters.
A set gives fast membership checks.

def count_vowels(s: str) -> int:
    vowels = set('aeiou')
    count = 0
    return count

3. Scan the String Once

Iterate through every character in the string.
If the lowercase character is in the vowel set, increment the counter.
The final counter is the answer.

def count_vowels(s: str) -> int:
    vowels = set('aeiou')
    count = 0
    for char in s:
        if char.lower() in vowels:
            count += 1
    return count

4. Handle Edge Cases Naturally

An empty string returns 0 because the loop never increments the count.
Strings with no vowels also return 0.
Uppercase-only inputs work because each character is converted with lower().

5. Final Solution and Complexity

The solution does one constant-time vowel check per character.
Time complexity is O(n), where n is the length of the string.
Space complexity is O(1) because the vowel set has a fixed size.

def count_vowels(s: str) -> int:
    vowels = set('aeiou')
    count = 0

    for char in s:
        if char.lower() in vowels:
            count += 1

    return count

Count Vowels – Solution & Complexity

1. Understand What Counts

2. Normalize Characters

3. Scan the String Once

4. Handle Edge Cases Naturally

5. Final Solution and Complexity

FAQ

What is the time and space complexity of Count Vowels?

What is the most efficient way to solve Count Vowels?

How do you solve Count Vowels in Python & JavaScript?

Is Count Vowels an easy or hard interview question?