What is the run time complexity of this for loop?

Question

I am trying to find out the run-time complexity of this algorithm.

public static void main(String[] args) throws InterruptedException{

  for (int N=100; N<=1000000; N=N*5) {  
   long start = System.currentTimeMillis();
   for (int i = 1; i <= N; i++) {     
      for (int j = 1; j <= Math.pow(N,1.5); j++) {
      i = i*2;
      j = j*2;
      Thread.sleep(10); 
     } 
    }
   long stop = System.currentTimeMillis();
   long elapsed = (long)(stop - start);
   System.out.println();
   System.out.println("For N=" + N + " RT in msec: "+elapsed); 
 }
}

The first for loop:

for (int N=100; N<=1000000; N=N*5) // runs n/5 times, so O(n). 

The first inner loop:

for (int i = 1; i <= N; i++) // runs n times. 

The second inner loop:

for (int j = 1; j <= Math.pow(N,1.5); j++) { // we can consider Math.pow O(1)
      i = i*2;
      j = j*2;
      Thread.sleep(10); 
     } 

So by multiplying all O(n) * O(n) * O(1) = O(n^2) Is my answer correct? I am a little confused on this. Will appreciate any clarification on this. Thank You

Answer 1

The first loop is actually O(k) which 5^k = N . Hence, k = log_5(N) . The first inner loop is true (in O(n) ). And the second inner loop j is each time is times to 2 . Hence, it is O(h) which 2^h = N^1.5 . Therefore, h = 1.5 log(N) .

In sum, the algorithm is in O(log_5(N) * N * log(N)) = O(N log^2(N)) .