[victorhn]

Naive solution: There are 924x1080 px as possible candidates for the upper-left corner of the new rectangle. Test each of these candidates to see if a rectangle placed there is a valid position. The complexity would be O(W * H * n) where W,H = dimensions in pixels of the screen, n = number of already placed windows. For practical purposes i think this would be a good solution, assuming the need for a new window is not very frequent. I suspect the complexity can be improved though, maybe with some sweep line algorithm.