What is the complexity of this code whose nested for loop repeatedly doubles its counter?

Question

In the book Programming Interviews Exposed it says that the complexity of the program below is O(N), but I don\'t understand how this is possible. Can someone expla

Answer 1

@Daniel Fischer's answer is correct.

I would like to add the fact that this algorithm's exact running time is as follows:

Which means:

Answer 2

You need a bit of math to see that. The inner loop iterates Θ(1 + log [N/(i+1)]) times (the 1 + is necessary since for i >= N/2, [N/(i+1)] = 1 and the logarithm is 0, yet the loop iterates once). j takes the values (i+1)*2^k until it is at least as large as N, and

(i+1)*2^k >= N <=> 2^k >= N/(i+1) <=> k >= log_2 (N/(i+1))

using mathematical division. So the update j *= 2 is called ceiling(log_2 (N/(i+1))) times and the condition is checked 1 + ceiling(log_2 (N/(i+1))) times. Thus we can write the total work

N-1                                   N
 ∑ (1 + log (N/(i+1)) = N + N*log N - ∑ log j
i=0                                  j=1
                      = N + N*log N - log N!

Now, Stirling's formula tells us

log N! = N*log N - N + O(log N)

so we find the total work done is indeed O(N).

Answer 3

Outer loop runs n times. Now it all depends on the inner loop.
The inner loop now is the tricky one.

Lets follow:

i=0 --> j=1             ---> log(n) iterations
...
...
i=(n/2)-1 --> j=n/2     ---> 1 iteration.
i=(n/2) -->   j=(n/2)+1 --->1 iteration.

---

i > (n/2)            ---> 1 iteration
(n/2)-1 >= i > (n/4) ---> 2 iterations
(n/4) >= i > (n/8)   ---> 3 iterations
(n/8) >= i > (n/16)  ---> 4 iterations   
(n/16) >= i > (n/32) ---> 5 iterations

(n/2)*1 + (n/4)*2 + (n/8)*3 + (n/16)*4 + ... + [n/(2^i)]*i

   N-1                                   
 n*∑ [i/(2^i)] =< 2*n
   i=0  

--> O(n)