I am using this solution to find rectangles aligned with the image border in a binary matrix. Suppose now I want to find a rectangle that is not aligned with the image border, a
I only have a partial answer for this question, and only a few thoughts on complexity or speed for what I propose.
Brute Force
The first idea that I see is to use the fact that your problem is discrete to implement a rotation around the center of the image and repeat the algorithm you already use in order to find the axis aligned solution.
This has the downside of checking a whole lot of candidate rotations. However, this check can be done in parallel since they are indepedant of one another. This is still probably very slow, although implementing it (shouldn't be too hard) and would provide a more definite answer to the question speed once parallelized.
Note that your work-space being a discrete matrix, there is only a finite number of rotation to browse through.
Other Approach
The second solution I see is:
This second solution would probably give you an approximation of the soluiton, but I believe it might prove to be worth trying.
For reference
The only solutions that I have found for the problem of the maximum/largest empty rectangle are axis-aligned. I have seen many unanswered questions corresponding to the oriented version of this problem on 2D continuous space.
EDIT:
[1] Since what we want is to separate the connected component, if there is a degree of overlap, you should do as in the following example:
0 1 0 0
0 1 0 1
0 0 0 1
should be divided into:
0 0 0 0
0 0 0 1
0 0 0 1
and
0 1 0 0
0 1 0 0
0 0 0 0
Note that I kept the original dimensions of the matrix. I did that because I'm guessing from your post it has some importance and that a rectangle expanding further away from the boundaries would not be found as a solution (i.e. that we can't just assume there are zero values beyond the border).
EDIT #2:
The choice of whether or not to keep the matrix dimensions is debatable since it will not directly influence the algorithm.
However, it is worth noting that if the matrices corresponding to connected components do not overlap on non-zero values, you may choose to store those matrices "in-place".
You also need to consider the fact that if you wish to return as output the coordinates of the rectangle, creating a matrix with different dimensions for each connected component, this will force you to store the coordinates of your newly created matrix in the original one (actually, one point, say for instance the up-left one, should be enough).