O(n^2) vs O (n(logn)^2)

Question

Is time complexity O(n^2) or O (n(logn)^2) better?

I know that when we simplify it, it becomes

O(n) vs O((logn)^2)

Answer 1

n is only less than (log n)² for values of n less than 0.49...

So in general (log n)² is better for large n...

But since these O(something)-notations always leave out constant factors, in your case it might not be possible to say for sure which algorithm is better...

Here's a graph:

(The blue line is n and the green line is (log n)²)

Notice, how the difference for small values of n isn't so big and might easily be dwarfed by the constant factors not included in the Big-O notation.

But for large n, (log n)² wins hands down: