Question

Ive been working with the running times of algorithms.

Im trying to determine the cost for n pushes onto a stack when the stack is full. That is when a given stack of size N is full replace it with a stack of size f(N) where f(N) = N + c (a constant) or f(N) = 2N.
I am to compare these 2 strategies.
you can assume that a regular push is constant time but a "special push" (when the stack is full) costs f(N) + N + 1 time. That is, we assume a cost of f(N) units to create the new
array, N units of time to copy the N elements and one unit of time to copy the new
element. Whats bigOh?

Answer 1

I believe the the first case, f(N) = N + c should run in O(n^2/c) but how do you get that answer?
f(N) + N +1 = time <=> N +c + N + 1 = time <=> 2N + k = time where k = c+1

Answer 2

The big O in the worst case is just \(f(n)\in O(n)\).