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?

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).

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.

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".

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.

Let m be the number of iterations taken. Then,

i+m = n^2 - m

which gives,

m = (n^2-i)/2

In Big-O notation, this implies a complexity of O(n^2).

Yes, this algorithm is O(n^2).

To calculate complexity, we have a table the complexities:

O(1) O(log n) O(n) O(n log n) O(n²) O(n^a) O(a^n) O(n!)

Each row represent a set of algorithms. A set of algorithms that is in O(1), too it is in O(n), and O(n^2), etc. But not at reverse. So, your algorithm realize n*n/2 sentences.

O(n) < O(nlogn) < O(n*n/2) < O(n²)

So, the set of algorithms that include the complexity of your algorithm, is O(n²), because O(n) and O(nlogn) are smaller.

For example: To n = 100, sum = 5000. => 100 O(n) < 200 O(n·logn) < 5000 (n*n/2) < 10000(n^2)

I'm sorry for my english.

> 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?

Yes! That's why it's O(n^2). By the same logic, it's a lot less than n * n * n, which makes it O(n^3). It's even O(6^n), by similar logic.

big-O gives you information about upper bounds.

I believe you are trying to ask why the complexity is theta(n) or omega(n), but if you're just trying to understand what big-O is, you really need to understand that it gives upper bounds on functions first and foremost.