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

You will have exactly n*n/2 loop iterations (or (n*n-1)/2 if n is odd). In the big O notation we have O((n*n-1)/2) = O(n*n/2) = O(n*n) because constant factors "don't count".