Question

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Big O when adding together different routines

What does O(n) + O(log(n)) reduce to? My guess is O(n) but can not give a rigorous reasoning.

I understand O(n) + O(1) should reduce to O(n) since O(1) is just a constant.

Answer 1

Well since O( f(n) ) + O( g(n) ) = O ( f(n) + g(n) ) We are simply trying to calculate an f(n) such that f(n) > n + log(n)

Since as n grows sufficiently log(n) < n we can say that f(n) > 2n > n + log(n)

Therefore O(f(n)) = O(2n) = O(n)

In a more general sense, O( f(n) ) + O( g(n) ) = O( f(n) ) if c*f(n)>g(n) for some constant c. Why? Because in this case f(n) will “dominate” our algorithm and dictate its time complexity.