@atul: given a matrix just like above, (usually an image) the pixel values with similar can be searched for around the current pixel, and they all can be marked in one go,
think of an algorithm, which does the following 1) when a one is replaced by '2' manually, then algorithm changes every '1' that is adjacent to '2' in a recursive fashion. 2) now, how many times the manual involvement is needed. assuming the algorithm spreads across diagonals too, then the below example has only one island, or as per my example, changing any one 1 to 2, will do the trick, 1100 1100 0011 --> 1 islands, or 1 manual changes 1100 1100 0001 --> 2 islands, or 2 manual changes 1101 1101 1100 0001 --> 3 islands, or 3 manual changes (if you didn't understand what a pixel filler is.. you can leave that for now.. its just one application of BFS) -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/DLCoXlVhNccJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
