I have 2 search algorithms and I have derived the following tight bound representations:
$$nlog (n) + mlog (n) $$ $$m*n $$ Now i want to find a function $f (n) $ so that when 'm' is an element of tight bound (f (n)), both algorithms have equal asymptotic run time.
Now I'm not sure if its just as simple as setting the two equations equal to each other and isolating 'm' or of it is more than that.
If you assume that $m = \Theta(\log n)$, then time complexities for your two algorithms will be: $$T_1(n) = \Theta(n \log n) + \Theta(\log^2 n) = \Theta(n \log n)$$ $$T_2(n) = \Theta(n \log n)$$