L2 · Working engineerPython internals~8 min · 3 tests#68

Decorator With Arguments (@repeat(n))

Write a parameterised Python decorator @repeat(n) that calls a function n times — three levels of nested functions explained, with live tests.

The problem

Write a decorator factory repeat(times) so that

@repeat(3)
def hello():
    print("Hello")

hello()

prints Hello three times. The wrapped function should return the result of its last call and keep its __name__.

Examples

Example 1

Input

captured(say_hello, repeat)

Expected output

'Hello\nHello\nHello\n'

Example 2
Input
```
returns_and_name(repeat)
```
Expected output
```
(10, 'double')
```

+ 1 hidden test on Submit.

Edge cases to ask about

times = 1
Return value
Name preserved

How the tests call your code

These helpers run before your code. The test inputs above call them.

def say_hello(repeat):
    @repeat(3)
    def hello():
        print("Hello")
    hello()

def returns_and_name(repeat):
    @repeat(2)
    def double(x):
        return x * 2
    return (double(5), double.__name__)

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 · repeat

import functools

def repeat(times):
    pass

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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
Three nested functions	O(times)	O(1)	factory(args) → decorator(func) → wrapper(*args).

Walkthrough of the optimal approach (try it yourself first)

@repeat(3) means hello = repeat(3)(hello). So repeat(times) must return a decorator, which takes func and returns a wrapper. Three nested functions, each closing over the variable it needs.

Keep functools.wraps on the innermost wrapper and return the last result.

Complexity: O(times) time, O(1) space. The wrapped function is called `times` times.

Reveal the reference solution

import functools

def repeat(times):
    def decorator(func):
        @functools.wraps(func)
        def wrapper(*args, **kwargs):
            result = None
            for _ in range(times):
                result = func(*args, **kwargs)
            return result
        return wrapper
    return decorator

Follow-ups interviewers ask

Make @repeat work with or without parentheses.
Write it as a class-based decorator.

Frequently asked interview questions

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

What is the time complexity of Decorator With Arguments (@repeat(n)) in Python?

The optimal solution runs in O(times) time and O(1) auxiliary space. The wrapped function is called times times.

What follow-up questions do interviewers ask about Decorator With Arguments (@repeat(n))?

Make @repeat work with or without parentheses. Write it as a class-based decorator.