algorithm - Running time complexity of double for-loops

Question

I'm somewhat confused by the following algorithms. In particular, I don't understand why the first is O(n) and the second is O(n^2). My only intuition is perhaps that the inner and outer loops for the first algorithm aren't "linked." Secondly, I can intuitively see that the second algorithm is O(n^2), but how would we go about finding some constants c1, c2 to prove f(n) is big-0 and little-0 of n^2?

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

---

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

Answer 1

To formally deduce the order of growth, you may proceed like the following:

For the c' case, the inner loop won't execute when i = 0, therefore, the solution is to externalize the outer loop "iterations" that won't affect the inner loop.