Question

given an n*m matrix with the possible values of 1, 2 and null:

  . . . . . 1 . .
  . 1 . . . . . 1
  . . . 2 . . . .
  . . . . 2 . . .
  1 . . . . . 1 .
  . . . . . . . .
  . . 1 . . 2 . .
  2 . . . . . . 1

I am looking for all blocks B (containing all values between (x0,y0) and (x1,y1)) that:

• contain at least one ‘1’
• contain no ‘2’
• are not a subset of a another block with the above properties

Example:

The red, green and blue area all contain an ‘1’, no ‘2’, and are not part of a larger area.
There are of course more than 3 such blocks in this picture. I want to find all these blocks.

what would be a fast way to find all these areas?

I have a working brute-force solution, iterating over all possible rectangles, checking if they fulfill the first two criteria; then iterating over all found rectangles, removing all rectangles that are contained in another rectangle; and I can speed that up by first removing consecutive identical rows and columns. But I am fairly certain that there is a much faster way.

Answer 1

I finally found a solution that works almost in linear time (there is a small factor depending on the number of found areas). I think this is the fastest possible solution.

Inspired by this answer: https://stackoverflow.com/a/7353193/145999 (pictures also taken from there)

First, I go trought the matrix by column, creating a new matrix M1 measuring the number of steps to the last ‘1’ and a matrix M2 measuring the number of steps to the last ‘2’

imagine a ‘1’ or ‘2’ in any of the grey blocks in the above picture

in the end I have M1 and M2 looking like this:

No go through M1 and M2 in reverse, by row:

I execute the following algorithm:

 foundAreas = new list()

 For each row y backwards:
     potentialAreas = new List()
     for each column x:
        if M2[x,y]>M2[x-1,y]:
            add to potentialAreas: 
                new Area(left=x,height=M2[x,y],hasOne=M1[x,y],hasTop=false)
        if M2[x,y]<M2[x-1,y]:
            for each area in potentialAreas:
                 if area.hasTop and area.hasOne<area.height:
                        add to foundAreas:
                             new Box(Area.left,y-area.height,x,y)
            if M2[x,y]==0: delete all potentialAreas
            else:
                 find the area in potentialAreas with height=M2[x,y] 
                 or the one with the closest bigger height: set its height to M2[x,y]
                 delete all potentialAreas with a bigger height

            for each area in potentialAreas:
                 if area.hasOne>M1[x,y]: area.hasOne=M1[x,y]
                 if M2[x,y+1]==0: area.hasTop=true

now foundAreas contains all rectangles with the desired properties.