Write a Python @retry decorator that re-runs a failing function up to N times, logs each attempt and re-raises after the last. Tested with a flaky function.
The problem
Write a decorator factory retry(max_attempts):
- call the function; if it raises, print
Attempt <k> -> Failedand try again - on success print
Attempt <k> -> Successand return the result - after
max_attemptsfailures, re-raise the last exception
A function that fails twice and then succeeds should print:
Attempt 1 -> Failed
Attempt 2 -> Failed
Attempt 3 -> Success
Examples
Example 1
Input
captured(run_retry, retry, 2, 3)
Expected output
'Attempt 1 -> Failed\nAttempt 2 -> Failed\nAttempt 3 -> Success\n'
Example 2
Input
run_retry(retry, 1, 3)
Expected output
'ok'
+ 2 hidden tests on Submit — re-raises after the last attempt.
Edge cases to ask about
- Success on first try
- All attempts fail
- Exception type preserved
How the tests call your code
These helpers run before your code. The test inputs above call them.
def flaky(fail_times, value="ok"): state = {"calls": 0} def call(): state["calls"] += 1 if state["calls"] <= fail_times: raise ConnectionError("temporary failure") return value return call def run_retry(retry, fail_times, max_attempts): fn = retry(max_attempts)(flaky(fail_times)) return fn() def exhausted(retry): fn = retry(3)(flaky(5)) out = captured(lambda: raises(fn)) return out, raises(retry(3)(flaky(5)))
Hints
0/3How an interviewer scores this
0/9Your 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.
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 |
|---|---|---|---|
| Loop inside the wrapper | O(k) | O(1) | Production versions add exponential backoff with jitter. |
Walkthrough of the optimal approach (try it yourself first)
Loop over attempts. try the call; except Exception prints the failure and, on the last attempt, re-raises with a bare raise (keeping the original traceback). The else branch runs only on success.
In production: catch only retryable exceptions (ConnectionError, timeouts — never KeyboardInterrupt), and sleep with exponential backoff and jitter between attempts so a thousand clients do not retry in lockstep. Libraries like tenacity do this.
Complexity: O(k) time, O(1) space. At most k = max_attempts calls of the wrapped function.
Reveal the reference solution
import functools def retry(max_attempts): def decorator(func): @functools.wraps(func) def wrapper(*args, **kwargs): for attempt in range(1, max_attempts + 1): try: result = func(*args, **kwargs) except Exception: print(f"Attempt {attempt} -> Failed") if attempt == max_attempts: raise else: print(f"Attempt {attempt} -> Success") return result return wrapper return decorator
Follow-ups interviewers ask
- Add exponential backoff with jitter.
- Retry only on specific exception types.
Frequently asked interview questions
Core interview concepts, complexities, and follow-ups scored by hiring teams.
What is the time complexity of Retry Decorator With Max Attempts in Python?
The optimal solution runs in O(k) time and O(1) auxiliary space. At most k = max_attempts calls of the wrapped function.
What follow-up questions do interviewers ask about Retry Decorator With Max Attempts?
Add exponential backoff with jitter. Retry only on specific exception types.
