Question

I am taking up algorithm course on coursera,and I am stuck on this particular problem. I am supposed to find the time complexity of this code.

int sum = 0

 for (int i = 1; i <= N*N; i = i*2)

  {
     for (int j = 0; j < i; j++)
           sum++; 
  }

I checked it in eclipse itself, for any value of N the number of times sum statement is executed is less than N

final value of sum:
for N=8 sum=3 
for N=16 sum=7 
for N=100000 sum=511

so the time complexity should be less than N
but the answer that is given is N raised to the power 2, How is it possible?

What I have done so far:

the first loop will run log(N^ 2) times, and consequently second loop will be execute 1,2,3.. 2 logN

Answer 1

The first inner loop will be 1 + 2 + 4 + 8 .. 2^M where 2^M is <= N * N.

The sum of powers of 2 up to N * N is approximately 2 * N * N or O(N ^ 2)

Note: When N=100000 , N*N will overflow so its result is misleading. If you consider overflow to be part of the problem, the sum is pretty random for large numbers so you could argue its O(1), i.e. if N=2^15, N^2 = 2^30 and the sum will be Integer.MAX_VALUE. No higher value of N will give a higher sum.