Solve the Easy-level Count Vowels coding interview problem online in Python & JavaScript. Run your code against real test cases in the browser and study worked examples. It covers strings. Read the step-by-step solution and complexity analysis. Java starter code is also included for reference (no in-browser JVM yet, so it can't be run here).

Count Vowels – Solve Online in Python & JavaScript

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Count ASCII vowels in a string without changing other characters.

Return the number of characters in `s` that are ASCII vowels (`a`, `e`, `i`, `o`, `u`) in either case.

**Input / output**
- Input: `s: string`
- Output: integer count

**Examples**
1. `"hello"` returns `2`.
2. `"AEIOU"` returns `5`.

**Constraints**
- `0 <= s.length <= 100,000`
- Only the ten ASCII vowel characters count

**Follow-up**
How would locale-aware vowel classification change the contract and tests?

- 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.

Understand What Counts

- 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.

Normalize Characters

- 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.

Scan the String Once

- 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()`.

Handle Edge Cases Naturally

- 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.