L1 · FoundationsText basics~2 min · 4 tests

Count the Vowels in a String

Count the vowels in a Python string, upper and lower case, with a loop and a set lookup in O(n). A beginner interview warm-up you can run in the browser.

The problem

Return how many vowels (a e i o u, either case) are in s.

Examples

Example 1
Input
```
count_vowels('Python Interview')
```
Expected output
```
5
```
Example 2
Input
```
count_vowels('rhythm')
```
Expected output
```
0
```

+ 2 hidden tests on Submit.

Edge cases to ask about

Empty string
Upper case
No vowels

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · count_vowels

def count_vowels(s):
    """Number of vowels (a, e, i, o, u, any case) in s."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

Your code runs in real CPython inside your browser — nothing is sent anywhere. The first run downloads the interpreter (about 6 MB, once). Your code is saved on this device as you type.

Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

Measure it

Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Loop + set membership	O(n)	O(1)	The vowel set has a fixed size.

Walkthrough of the optimal approach (try it yourself first)

Walk the string and count characters found in a vowel set. sum(1 for ch in s if ch in vowels) does it in one line without building a list. Lower-casing first (s.lower()) works too but creates a copy.

Complexity: O(n) time, O(1) space. Each character is checked once against a fixed 10-letter set.

Reveal the reference solution

def count_vowels(s):
    vowels = set("aeiouAEIOU")
    return sum(1 for ch in s if ch in vowels)

Follow-ups interviewers ask

Return counts per vowel.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Count the Vowels in a String in Python?

The optimal solution runs in O(n) time and O(1) auxiliary space. Each character is checked once against a fixed 10-letter set.

What follow-up questions do interviewers ask about Count the Vowels in a String?

Return counts per vowel.