Improving O(n) while looping through a 2d array in C++
An answer to this question on Stack Overflow.
Question
A goal of mine is to reduce my O(n^2) algorithms into O(n), as it's a common algorithm in my Array2D
As you can see, I reduced my doubly-nested for loops into a single for loop here. It's running fine when I execute it. Speed has surely improved. Is there any other way to improve the speed of this member function? I'm hoping to use this algorithm as a model for my other member functions that have similar operations on multidimensional arrays.
/// <summary>
/// Fills all items within the array with a value.
/// </summary>
/// <param name="ob">The object to insert.</param>
void fill(const T &ob)
{
if (m_array == NULL)
return;
//for (int y = 0; y < m_height; y++)
//{
// for (int x = 0; x < m_width; x++)
// {
// get(x, y) = ob;
// }
//}
int size = m_width * m_height;
int y = 0;
int x = 0;
for (int i = 0; i < size; i++)
{
get(x, y) = ob;
x++;
if (x >= m_width)
{
x = 0;
y++;
}
}
}
Answer
Make sure things are contiguous in memory as cache behavior is likely to dominate the run-time of any code which performs only simple operations.
For instance, don't use this:
int* a[10];
for(int i=0;i<10;i++)
a[i] = new int[10];
//Also not this
std::vector< std::vector<int> > a(std::vector<int>(10),10);
Use this:
int a[100];
//or
std::vector<int> a(100);
Now, if you need 2D access use:
for(int y=0;y<HEIGHT;y++)
for(int x=0;x<WIDTH;x++)
a[y*WIDTH+x];
Use 1D accesses for tight loops, whole-array operations which don't rely on knowledge of neighbours, or for situations where you need to store indices:
for(int i=0;i<HEIGHT*WIDTH;i++)
a[i];
Note that in the above two loops the number of items touched is HEIGHT*WIDTH in both cases. Though it may appear that one has a time complexity of O(N^2) and the other O(n), it should be obvious that the net amount of work done is HEIGHT*WIDTH in both cases. It is better to think of N as the total number of items touched by an operation, rather than a property of the way in which they are touched.