L1 · FoundationsText basics~2 min · 5 tests

Count the Words in a String

Count the words in a string correctly even with leading, trailing and repeated spaces, using split(). A quick Python string-handling interview check.

The problem

Return the number of words in s, where words are separated by any amount of whitespace.

Examples

Example 1
Input
```
count_words('Python is fun')
```
Expected output
```
3
```
Example 2
Input
```
count_words('  many   spaces here ')
```
Expected output
```
3
```

+ 3 hidden tests on Submit — tabs and newlines count as whitespace.

Edge cases to ask about

Only spaces
Tabs/newlines

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_words

def count_words(s):
    """Number of whitespace-separated words."""
    # 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 read why.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

From brute force to optimal

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

Approach	Time	Space	Idea
len(s.split())	O(n)	O(n)	A manual state machine counting word starts uses O(1) space.

Walkthrough of the optimal approach (try it yourself first)

len(s.split()). With no argument, split() treats any run of spaces, tabs or newlines as one separator and ignores the ends. s.split(" ") would count empty strings between double spaces.

For O(1) extra space, count positions where a non-space follows a space (or the start).

Complexity: O(n) time, O(n) space. split() scans the string once and stores the words.

Reveal the reference solution

def count_words(s):
    return len(s.split())

Follow-ups interviewers ask

Count words without building a list.

Frequently asked interview questions

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

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

The optimal solution runs in O(n) time and O(n) auxiliary space. split() scans the string once and stores the words.

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

Count words without building a list.