Why is the Big-O complexity of this algorithm O(n^2)?

I know the big-O complexity of this algorithm is O(n^2), but I cannot understand why.

int sum = 0; 
int i = 1; j = n * n; 
while (i++ < j--) 
  sum++;

Even though we set j = n * n at the beginning, we increment i and decrement j during each iteration, so shouldn't the resulting number of iterations be a lot less than n*n?

Solution 1:

During every iteration you increment i and decrement j which is equivalent to just incrementing i by 2. Therefore, total number of iterations is n^2 / 2 and that is still O(n^2).

Solution 2:

big-O complexity ignores coefficients. For example: O(n), O(2n), and O(1000n) are all the same O(n) running time. Likewise, O(n^2) and O(0.5n^2) are both O(n^2) running time.

In your situation, you're essentially incrementing your loop counter by 2 each time through your loop (since j-- has the same effect as i++). So your running time is O(0.5n^2), but that's the same as O(n^2) when you remove the coefficient.