[lesuorac]

This is typically why we don't pick N to mean all of the resources in the universe. I guess you're the kind of guy that chooses to uses not 8-bit bytes in an interview?

Instead typically it's the amount of accesses into some array. Other times people get cute and it's `log2(input length)` so somebodies algorithm could be O(N) while if N was just `input length` then it'd be N^2.

[Dylan16807]

This isn't about picking N. This is saying "No matter what N you pick, your runtime will be smaller than roughly 2^[number of atoms in the reachable universe]" (Because if you hit the same state twice, you're stuck in a loop that will never halt.)

So yes, your algorithm is bound by O(N^2), and that's the number that matters on practical computers... but it's also bound by O(1) with a stupidly large constant. So it's O(1).

> I guess you're the kind of guy that chooses to uses not 8-bit bytes in an interview?

Dude, I'm here specifically to argue against insisting on the exact literal meaning, and to choose the practical option.