I don't understand this task at all.

Actually, a point to be made, in Java at least, it is set up as y,x coordinates. The points of the graph are: [0,0],[0,1],[0,2],[0,3] etc... [1,0],[1,1],[1,2],[1,3] etc... [2,0],[2,1],[2,2],[2,3] etc.. etc..,etc..,etc..,etc... So in the starting code, you could output a value like this:

System.out.print(a1[3,1]);
// 4 actual rows down, 2 actual columns over (counts start at 0)

which would output 0,

Before I asked the question, I plotted all these out on a graph. The written out 2D array is the reflection (on the x plane) of the plotted graph. What stumped me is that a single point counts as a rectangle which was what was throwing me off originally.

This task was easier than the Armstrong numbers task. That one took me the better part of 2 weeks, and i recruited 2 friends who are professional programmers for help, but ya, I had to do the same thing with plotting things out... and the reverse x,y thing threw me off for a bit too.

My solution is really messy (I think). Could you PM me yours so I can compare and attempt to make mine more elegant?

Sure. I don't know if you have figured this out yet, but if delete the last curly bracket (the closing one fore the Solution class) then add it right back, IntelliJ will properly format you code, e.g. add space, new lines, indentation. Sometimes my code gets a little sloppy when I am first trying to solve something, with adding and deleting code, but I do this alot to quickly clean things up. Take this code for example:

private static int findMaxY(int x, int y, byte[][] a) {
    int count = 0;for (int i=y+1;i<a.length;i++) {
        if (a[i][x]==1)count++;else break; }

return count;
}

when i do what i described above it turns it into this:

private static int findMaxX(int x, int y, byte[][] a) {
    int count = 0;
    for (int i = x + 1; i < a[1].length; i++) {
        if (a[y][i] == 1) count++;
        else break;
    }
    return count;
}

This works for any closing curly bracket and the code contained within. I always use the last one to clean up all my code.

Cool tip, thanks!

so 0 0 0 1 count as a rectangle And 0 1 0 1 count as a rectangle. Am i right.

Please could you explai more how you get this for rectangle in your example ??? : 1,1,0,0,0,0,1,1,1 1,1,0,0,0,0,1,1,1 0,0,0,0,0,0,0,0,0 1,1,1,1,1,1,1,0,0 1,1,1,1,1,1,1,0,0 0,0,0,0,0,0,0,0,1 This has 4 rectangles in it. The key to this is that the rectangles do not touch or overlap, so there are 0's (or the edge) around them.