Big-O Time Complexity of nested for loops [duplicate]

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?

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

• 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 $\Theta$. – Raphael Mar 2 '15 at 0:03
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.
• Make sure that you understand the concept of running time. If a line is run $s$ times then the running time is at least $s$. – Yuval Filmus Mar 1 '15 at 22:22