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 explain why this is?

int var = 2;
for (int i = 0; i < N; i++) {
   for (int j = i+1; j < N; j *= 2) {
      var += var;
   }
}

Answer 1

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).