L2 · Working engineerStrings~8 min · 5 tests#26

String Compression (Run-Length Encoding)

Compress consecutive repeated characters as char+count, like 'aaabbcccc' → 'a3b2c4', in one O(n) pass. Practise run-length encoding in Python.

The problem

Compress s by replacing each run of the same character with the character followed by the run length.

"aaabbcccc" → "a3b2c4". Every run gets a count, even a run of 1 ("abc" → "a1b1c1"). Empty input gives "".

Examples

Example 1
Input
```
compress('aaabbcccc')
```
Expected output
```
'a3b2c4'
```
Example 2
Input
```
compress('abc')
```
Expected output
```
'a1b1c1'
```

+ 3 hidden tests on Submit — same char in two separate runs.

Edge cases to ask about

Empty string
Single character
Same character in separate runs
Run of 10+ (two-digit count)

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 · compress

def compress(s):
    """Run-length encode: 'aaab' -> 'a3b1'."""
    # 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
String += in a loop	O(n) amortised in CPython	O(n)	Works, but relies on an implementation detail.
List of parts + join	O(n)	O(n)	One pass, flush each run when the character changes.
bestitertools.groupby	O(n)	O(n)	''.join(f'{k}{len(list(g))}' for k, g in groupby(s)).

Walkthrough of the optimal approach (try it yourself first)

Walk the string keeping run_char and run_len. When the character changes, append f"{run_char}{run_len}" to a list and start a new run. After the loop, flush the last run — the most commonly forgotten line.

itertools.groupby groups consecutive equal items and makes this a one-liner.

Complexity: O(n) time, O(n) space. Single pass; the output can be up to 2n characters long.

Reveal the reference solution

def compress(s):
    if not s:
        return ""
    parts = []
    run_char, run_len = s[0], 1
    for ch in s[1:]:
        if ch == run_char:
            run_len += 1
        else:
            parts.append(f"{run_char}{run_len}")
            run_char, run_len = ch, 1
    parts.append(f"{run_char}{run_len}")
    return "".join(parts)

Follow-ups interviewers ask

Return the original if the compressed form is not shorter.
Write the decoder.

Frequently asked interview questions

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

What is the time complexity of String Compression (Run-Length Encoding) in Python?

The optimal solution runs in O(n) time and O(n) auxiliary space. Single pass; the output can be up to 2n characters long.

What is the brute-force approach, and how do you optimise it?

String += in a loop: O(n) amortised in CPython time, O(n) space. Works, but relies on an implementation detail. List of parts + join: O(n) time, O(n) space. One pass, flush each run when the character changes. itertools.groupby: O(n) time, O(n) space. ''.join(f'{k}{len(list(g))}' for k, g in groupby(s)).

What follow-up questions do interviewers ask about String Compression (Run-Length Encoding)?

Return the original if the compressed form is not shorter. Write the decoder.