Big O Notation and Time Complexity of C++ Code Snippet

Question

So I am looking for confirmation of what the time complexity of the c++ code snippet is:

for(int i = 0; i<N, i++){
    for(int k = 1; k<N; k*=2){
    //code with O(1)
    }
 }

I am thinking this would be O(NlgN) where lg is log base 2. The inner loop would be O(lgN) since k is doubling after each iteration. The outer loop is clearly O(N), making the whole code:

 O(N)*O(lgN) = O(NlgN).

Answer 1

Yes it is in O(n log n) but the base does not matter in big O notation since f=n \cdot log_2(n) \in \mathcal{O}(log_2(n) * n ) \subseteq \mathcal{O}(\frac{ln(n)}{ln(2)} * n ) \subseteq \mathcal{O}(log(n) * n ) \ni f = n \cdot ln (n) i.e.

Note the log at the end should still be ln, but people are not caring about the confusion to whenever log is to the base of 10 or e since it does not matter in big O.

So even for(int k = 2; k<N; k*= k) would be the same in complexity when using big O notation. However sometimes people are writing down the constant factors when comparing very minor optimisations, but thats not feasible unless you are talking about the quick sort implementation that runs on billions of instances around the world.

For the part how we can be sure that your inner loop is in deed bound by log(n) I kind of also did not find a nice math proof. Of course executing it is kind of a proof, but my theoretical approach is that, we can agree the inner loop executes as often as your function k *= 2 needs a bigger argument to reach n, so where k(x) >= n, and what do we know which x do we need to get the k(x) we want, is the inverse function k^(-1), and the inverse functions for 2^x is log_2(x).