Question

 for (int j=0,k=0; j<n; j++)
   for (double m=1; m<n; m*=2)
      k++;

I think it’s O(n^2) but I’m not certain. I’m working on a practice problem and I have the following choices:

• O(n^2)
• O(2^n)
• O(n!)
• O(n log(n))

Answer 1

Its O(nlog₂n). The code block runs n*log₂n times.

Suppose n=16; Then the first loop runs 16 (=n) times. And the second loops runs 4(=log₂n) times (m=1,2,4,8). So the inner statement k++ runs 64 times = (n*log₂n) times.