Which is better: O(n log n) or O(n^2)

Question

Okay so I have this project I have to do, but I just don\'t understand it. The thing is, I have 2 algorithms. O(n^2) and O(n*log₂n)

Answer 1

Big-O notation is a notation of asymptotic complexity. This means it calculates the complexity when N is arbitrarily large.

For small Ns, a lot of other factors come in. It's possible that an algorithm has O(n^2) loop iterations, but each iteration is very short, while another algorithm has O(n) iterations with very long iterations. With large Ns, the linear algorithm will be faster. With small Ns, the quadratic algorithm will be faster.

So, for small Ns, just measure the two and see which one is faster. No need to go into asymptotic complexity.

Incidentally, don't write the basis of the log. Big-O notation ignores constants - O(17 * N) is the same as O(N). Since log₂N is just ln N / ln 2, the basis of the logarithm is just another constant and is ignored.