Big O for 2D for loop

Question

I have to write down the Big O notation of an algorithm I had to think up for my homework.

I'm able to tell that the code below is O(n^2). Because for every x I have to go through all of the y's and it becomes slower as the world grows larger.

int[][] world = new world[20][20];
for (int x = 0; x < 20; x++)
{
    for (int y = 0; y < 20; y++)
    {
        ..
    }
}

But, for another question I have to go through the bottom half of the world, so my y loop gets halved.

int[][] world = new world[20][20];
for (int x = 0; x < 20; x++)
{
    for (int y = 10; y < 20; y++)
    {
        ..
    }
}

I'm not quite sure what Big O notation is appropriate for the above loop, is it still O(n^2) because it still becomes slower the bigger the world gets? Or is it O(log n) because the y is halved?

Answer 1

it is simply speaking O(n*n/2)=O(n^2) since constants arent considered in big O.

Answer 2

It's O(n^2) as y is still a function of n i.e. O(n^2/2) is still O(n^2).