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This question already has an answer here:

My gut tells me the time-complexity of the following code is simply O(n^2). However, I'm not convinced, thinking it could possibly be O(n^3):

cin >> n;
sum = 0;
for (int i = 0; i < n; i++)
    for (int j = 0; j < n * n; j++)
        sum++;

Can anyone provide a distinction, and why?

Thanks in advance.

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marked as duplicate by David Richerby, Raphael Mar 2 '15 at 0:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See also runtime-analysis+loops. And, important, here; $O(n^2)$ and $O(n^3)$ are not mutually exclusive, in fact $O(n^2) \subset O(n^3)$. You want to talk about $\Theta$. $\endgroup$ – Raphael Mar 2 '15 at 0:03
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Hint: What is the value of sum at the end of the code, as a function of $n$? If it is $s$ then the fifth line must have been run exactly $s$ times.

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  • $\begingroup$ So, I ran the program and input 5, and the value of sum was 5^3 = 75. Does this mean it is O(n^3) ? Still not convinced. $\endgroup$ – user29291 Mar 1 '15 at 22:20
  • $\begingroup$ Make sure that you understand the concept of running time. If a line is run $s$ times then the running time is at least $s$. $\endgroup$ – Yuval Filmus Mar 1 '15 at 22:22
  • $\begingroup$ Thank you. I mistyped in my previous comment, should be 125 instead of 75. $\endgroup$ – user29291 Mar 1 '15 at 22:29

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