Why is the Fibonacci Sequence Big O(2^n) instead of O(logn)?

Question

I took discrete math (in which I learned about master theorem, Big Theta/Omega/O) a while ago and I seem to have forgotten the difference between O(logn) and O(2^n) (not in the

Answer 1

With the recursive algo, you have approximately 2^N operations (additions) for fibonacci (N). Then it is O(2^N).

With a cache (memoization), you have approximately N operations, then it is O(N).

Algorithms with complexity O(N log N) are often a conjunction of iterate over every item (O(N)) , split recurse, and merge ... Split by 2 => you do log N recursions.