i don't need the code but suggestions, I'm stuck in this problem, actually I tried 4 different algoriths, each algorithm better than the previous one. (I won't tell you the algorithms because I think that doesn't help to other developers to learn).

What I can tell you is that my last code is pretty efficient and fast (not enough). I have some ideas but all of them are based on the idea of reducing the chances of using small words. For example... the string "ab" is contained in "abcd" so if the word is "abcde" then it is better to start with "abcd" instead of "ab" and then realizing that it was not the correct option. But I have always considered that this kind of improvements does not reduce the logarithm complexity, it just reduces some cycles of execution

That’s an interesting challenge! It looks like a fun way to test string manipulation. Speaking of breaking things down, it kind of reminds me of how we analyze content on social media. You can think of posts like "passwords" and your feed as an attempt to "crack" or curate based on what you follow or engage with. It’s a mix of fitting the right pieces together to get the perfect result, just like figuring out the correct password combinations!

I believe the main issue here is when we have passwords like this: [...,'aaaa' , 'aaaaaa',...]

and loginAttempt like this: ...aaaaaaaaaa...

We can find 'aaaa' + 'aaaa' but will remain 'aa' giving us a 'WRONG PASSWORD'. But, if we take 'aaaaaa' and then 'aaaa' we're gonna find a good answer.

return 'WRONG PASSWORD' if match == login: return res + match return passwordCracker, (passwords, login(minara.18_, res + match + ' ')

there are so many test cases and it seems working but most of them are failing because of time limit.

`python

for x in passwords: if x == loginAttempt[:len(x)]: # print(x) if x == loginAttempt: return x res = passwordCracker(passwords, loginAttempt[len(x):]) if res != 0 and 'WRONG PASSWORD' not in res: return x +' '+ res else: continue return 'WRONG PASSWORD'`