Nah, every algorithm that is O(N logN) is also O(n^2). That would be a sign, that you as an interviewer didn't fully grasp Big O.
That would be a sign that you as a candidate are being needlessly pedantic and usually a red flag.
Honestly it sounds as if you might not understand the difference between Theta and O. I understand that a function that is O(N^2) grows no faster than O(N^3) asymptotically speaking however the intention of my original comment and example is quite clear about their context.
Please stop, you are adding nothing to the discussion.
Maybe you might reread my original comment. This whole side discussion is of no value to my original comment which has a very clear and very narrow context.
I am not sure why the both of you want to belabor some ancillary talking point that you yourselves decided to introduce. It is of exactly no value to the context of my original comment and not in the least bit productive.
Also the fact that you're plugging numbers in for n indicates to me that you don't actually understand big O notation, which is surprising since the original article is decent.
[0] https://en.wikipedia.org/wiki/Big_O_notation#Formal_definiti...