I think you can also speed this up by incrementing i to the next candidate window end (increment by window_length - count). I.e. given "abcdefggqwertyu", if you

I think you can also speed this up by incrementing i to the next candidate window end (increment by window_length - count). I.e. given "abcdefggqwertyu", if your previous window was "[abcdefg]" and you see that "g" conflicted so you reset count to 0, the next possible window is "[gqwerty]". So you don't have go through the intermediate windows like "[efgqgq]" because you know that count can never be 4 within there. Edit: I think your solution already effectively does this, it's just that because you maintain invariant that the window always contains unique chars to avoid any additional datastructure, once the window start gets reset to "[g]" it needs to build back up the intermediate state one char a time to works it way back up to [gqwerty]. So the best case is approximately linear if count gets reset often, worst case is O(NW). The method I was thinking of directly tests the next window starting from its endpoint backwards, to do this you need a hashet. I think that would achieve on best case O(N/W) effectively sublinear, worst case is still O(NW) though.