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

Question

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--) 
  s         


        

Answer 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).