Question

Is the Big-O for the following code O(n) or O(log n)?

for (int i = 1; i < n; i*=2)
        sum++;

It looks like O(n) or am I missing this completely?

Answer 1

It is O(logn), since i is doubled each time. So at overall you need to iterate k times, until 2^k = n, and in this case it happens when k = logn (since 2^logn = n).

Simple example: Assume n = 100 – then:

iter1: i = 1
iter2: i = 2
iter3: i = 4
iter4: i = 8
iter5: i = 16
iter6: i = 32
iter7: i = 64
iter8: i = 128 > 100

It is easy to see that an iteration will be added when n is doubled, which is logarithmic behavior, while linear behavior is adding iterations for a constant increase of n.

P.S. (EDIT): mathematically speaking, the algorithm is indeed O(n) – since big-O notation gives asymptotic upper bound, and your algorithm runs asymptotically “faster” then O(n) – so it is indeed O(n) – but it is not a tight bound (It is not Theta(n)) and I doubt that is actually what you are looking for.