I think this is a tough kyu 5 ;)

to be honest, this should've been 8kyu

Ranks cannot be changed, but I totally agree, it's a bit too easy for a 6 kyu. A small bit of math would do the trick well.

do you mean something different from the worst case or what?

Yeah, I messed it up. I calculated the repeated string concatenation as T(n) = n^2 as in the worst case, but comparisons as T(n) = 1 as in "more realistic" scenario (i.e. you're unlikely to deal with GigaByte-sized strings, and the constant factor is so low, the O(n) comparsion can be negligible). This particular solution is indeed O(n^3)

What is n?

In the worst scenario all the inputs' sizes are inifinitely big, so both the string and the list will be n, no?

while it could be O(len(s)) by using a set

With hashing a newly created string - yes.

So the final algorithm is O(len(s) ** 2 * len(l))

Looks correct to me for the worst case.

O(len(l)) for creating the set

Only if the hashes are already calculated and cached; otherwise len(l[i]) has to be somewhere as well.

This solution is O(n^2)

What is n? And do you mean something different from the worst case or what? Arbitrary strings can't be compared in O(1).

So the final algorithm is O(len(s) ** 2 * len(l)) instead of the O(max(len(l), len(s) ** 2))

That's not how O works, and your reasoning is wrong too. This solution is O(n^2), and there's no asymptotically better algorithm for this task.

I did but this expression has many interpretations and I wanted to know yours.
Thanks for your kindness.

So here comes my first suggestion: Hankuna matata!

I don't understand... Say it in English:-)

Sorry, it is a suggestion, not an issue... Nevertheless I added it.
Please verify and stop with suggestions:-) CW is so slow that it takes times to re-publish and most often it times out when re-publishing.

Added though without any interest since everybody already knows the answer:-)

I find 5 kyu about right. Oh, and thanks for the compliment!

I did not look up the implementation of 'all' in cpython, but my tests show that all seems to end when one 'False' is found.
So no, i think this is pretty efficient, as he is passing an iterator not a list to all.

EDIT: Buildin all indeed breaks if it encounters a false, see: 1