public void m1(int n) {
   sum = 0;
    for (int i = n/2; i <= n; i++) {
        for (int j = 1; j <= n; j = 2 * j) {
            for (int k = 1; k <= n; k = k * 2) {

So, I know that the innermost loop and second loop are $\log(n)$ and the outermost loop is essentially $O(n)$, so that means $O(n\log(n)\log(n))$. But I am lost on what to do next. I am between $O(n^2 \log(n))$ and $O(n \log^2(n))$. Can someone tell me which is right and explain why?

  • $\begingroup$ So what is n (log n) (log n)? Don’t you trust yourself enough to figure this out? What about using a spreadsheet and evaluating all three expressions for n=1 to 100? $\endgroup$
    – gnasher729
    Feb 7 at 8:36
  • $\begingroup$ Both of them are right. $O(n \log^2 n) \subset O(n^2 \log n)$. Also, $O(f(n))$ denotes a set of functions so it does not make sense to say that a "a loop is $O(n)$". You are probably referring to the number of iterations of the loop. $\endgroup$
    – Steven
    Feb 7 at 10:20
  • 2
    $\begingroup$ The question is whether $n \log(n) \log(n) = n^2 \log(n)$ or $n \log(n) \log(n) = n \log^2(n)$. First, this question is off-topic for a computer science site. Second, you need to do your own homework before asking a math site, have you reviewed your high school math book or asked a friend? $\endgroup$ Feb 7 at 14:21

1 Answer 1


You are right, the two innermost loops perform $\Theta(\log n)$ iterations each, so we have a total of $\Theta(\log^2 n)$ iterations, which are repeated $\Theta(n)$ times in the outer loop, which implies that we have a total of $\Theta(n\log^2 n)$ iterations.

You just have to multiply the number of iterations of each loop since they are nested and all the iterations are performed without interruptions (we just increase a variable inside the loop). So we can not only give an upper-bound ($O(n\log^2 n)$ ) but also a lower-bound ($\Omega(n\log^2 n)$), which implies a complexity of $\Theta(n\log^2 n)$.


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