Part 1 and part 2 are the same, only the window size varies. But within a part, the window size is constant. For most, it should have just meant that, if you hard-coded the window size during part 1, that you need to make it a parameter.
But for both parts, the naïve solution should be O(N * W), where N is the input length, and W is the window size. Like a sibling says, since within a part the window size is fixed, it is acceptable to call it O(N).
Edit: ah, I see what his solution is doing. It is correct. I need to adjust my definition of naïve, I guess. His is O(N * W²). You can still reasonably consider W² constant, though, I think. And he gets to what I would call the "naïve" solution.
(Previously.) ~The "naïve solution" in the OP does not appear to be correct. Or at least, if it does work,~ it appears to do far more computation that it needs to.
The "xor trick" is O(N), too. (There cannot be a more efficient solution, as the entire input must be considered in the worst case. TFA is correct that its final form omits W, though.)
TFA's implementation is using a loop over the window size to compare each character to all the other characters, to see if they're unique. So, the article's O(N * W²) is correct, but I agree, the window size is fixed and small; it's fair to boil it down to just O(N).
(I never even considered doing a for loop like that to determine if the window is unique; I just histogram the window, and then check if all the values in the histogram are 1. If yes, unique. If I had been cleverer, I wouldn't rebuild the histogram as the window shifts, but there was no need for that. This is essentially the next solution TFA presents.)
