Is it possible to simplify the complexity $O(n\log{}n + \frac{n^2}{m}\log{}m)$?

Question

The question is in the title: is it possible to simplify the complexity $O(n\log{}n + \frac{n^2}{m}\log{}m)$ ? $n$ and $m$ are two variables, and you know that $n > m$ (by the way, what if we don't know that?).

What I first thought was that it is possible to reduce it by keeping only the second term: $O(\frac{n^2}{m}\log{}m)$, because the term $n^2$ is dominating the term $n$, even if we divide $n^2$ by $m$. But the more I think about it, the less confident I am about this.

Any idea/explanation? That would help a lot.

Thanks!

Answer 1

In general, if you have two parameters $m$ and $n$ that can be independently adjusted, you are stuck with two parameters. Sometimes $m$ and $n$ are naturally tied together (maybe they're roughly proportional in most cases, or maybe $m$ is generally $O(1)$ and its typical values don't depend much on $n$) and then you can reduce it.

For instance if you assume $m$ does not change and you are interested in the scaling of complexity with $n,$ then your approach of fixing $m$ and regognizing that the second term scales faster than the first would be valid.

However, all you are told is that $m

All of this assumes that $m$ is (loosely) a function of $n$ in the sense that typical values of $m$ depend on $n$ in a predictable way. As I said at the beginning, if there's no such relationship and the $m$ and $n$ are pretty much independent, two parameters means two parameters.