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

Your algorithm is equivalent to

while (i += 2 < n*n) 
  ...

which is O(n^2/2) which is the same to O(n^2) because big O complexity does not care about constants.