Quick Running Time Question I have a quick question about some running time stuff. In my algorithm I run merge sort twice, then loop $n$ times. If this is the case, does this make sense? $\Theta(nlogn+nlogn+n) = \Theta(nlogn)$. I am a bit uncertain about the $\Theta(nlogn+n) = \Theta(nlogn)$ aspect but fairly certain this is a true statement. Edit : Should add that only constant operations occur within the loop.

Yes, it's true. Since $n \log{n} > n$, complexity grows (is dominated by) $n \log{n}$.