L2 · Working engineerRecursion~10 min · 4 tests#52

Generate All Permutations of a String

Generate every permutation of a string with recursive backtracking in O(n·n!). Compare with itertools.permutations and test it in your browser.

The problem

Return a list of every permutation of s (assume its characters are distinct). Order of the list does not matter. Do not use itertools.

Examples

Example 1

Input

permutations('abc')

Expected output

['abc', 'acb', 'bac', 'bca', 'cab', 'cba']

Example 2
Input
```
permutations('ab')
```
Expected output
```
['ab', 'ba']
```

+ 2 hidden tests on Submit — 24 results.

Edge cases to ask about

Single character
Duplicate characters (follow-up)

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 · permutations

def permutations(s):
    """All permutations of s (distinct characters)."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

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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
Pick each character first, permute the rest	O(n · n!)	O(n · n!)	n! results, each n characters long.

Walkthrough of the optimal approach (try it yourself first)

For each character, fix it as the first and append it to every permutation of the remaining characters. The base case is length ≤ 1.

The output alone has n! entries, so nothing can beat O(n · n!) — the interviewer is checking you know that factorial growth is unavoidable here.

Complexity: O(n · n!) time, O(n · n!) space. There are n! permutations and building each costs O(n); storing them all needs the same.

Reveal the reference solution

def permutations(s):
    if len(s) <= 1:
        return [s]
    out = []
    for i, ch in enumerate(s):
        for rest in permutations(s[:i] + s[i + 1:]):
            out.append(ch + rest)
    return out

Follow-ups interviewers ask

Handle duplicate characters without duplicate results.
Yield them lazily with a generator.

Frequently asked interview questions

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

What is the time complexity of Generate All Permutations of a String in Python?

The optimal solution runs in O(n · n!) time and O(n · n!) auxiliary space. There are n! permutations and building each costs O(n); storing them all needs the same.

What follow-up questions do interviewers ask about Generate All Permutations of a String?

Handle duplicate characters without duplicate results. Yield them lazily with a generator.