at job interview

“ah sure, I’ll solve it in n log(n) for you with an obscure algorithm because n² is too slow for your 1000 customers, even though there’s no perceived difference for n<10⁷.”

Acshually, in the context of O(N^2) N can be seen to constantly be equal to N and thus, as a constant, we can ignore it in our O analysis.

Yes, my bubble sort does run in O(1)

O(n!^n!)

It works really well, until n=3, which takes a while. Don’t ask about n=4.