算法的时间复杂度 - n或n * n？

Question

I'm trying to find out which is the Theta complexity of this algorithm. (a is a list of integers)

def sttr(a):
    for i in xrange(0,len(a)):
        while s!=[] and a[i]>=a[s[-1]]:
            s.pop()
        s.append(i)
    return s

On the one hand, I can say that append is being executed n (length of a array) times, so pop too and the last thing I should consider is the while condition which could be executed probably 2n times at most.

From this I can say that this algorithm is at most 4*n so it is THETA(n).

But isn't it amortised analysis?

On the other hand I can say this:

There are 2 nested cycles. The for cycle is being executed exactly n times. The while cycle could be executed at most n times since I have to remove item in each iteration. So the complexity is THETA(n*n).

I want to compute THETA but don't know which of these two options is correct. Could you give me advice?

Answer 1

The answer is THETA(n) and your arguments are correct.

This is not amortized analysis.

To get to amortized analysis you have to look at the inner loop. You can't easily say how fast the while will execute if you ignore the rest of the algorithm. Naive approach would be O(N) and that's correct since that's the maximum number of iterations. However, since we know that the total number of executions is O(N) (your argument) and that this will be executed N time we can say that the complexity of the inner loop is O(1) amortized.