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area will be fixed to be 1×1 unit cell.
As for the numbers of times that we repeated the process,
after we generated the rectangles, we removed the small
corner rectangle, then generated the cells according to
the the zig-zag and then the parallelogram, and at last,
we generated the lower left rectangle, and so on. For each
time we re-generate, we added 1 to the number count variable.
I believe that this is not the most efficient way to do it.
Can you come up with a more efficient way? Thanks!

A:

If I understand what you are asking correctly, you want to generate a $1\times 1$ square, $1 \times 3$ rectangle, $2 \times 1$ rectangle, $2 \times 3$ rectangle, and so on. For each square, you want a corner removed, and then you want to generate a new 1×1 square.
The time taken by the method of your previous question is $O(n)$, since you need to generate the squares and rectangles at each step. The algorithm you are trying to implement is already doing the same thing. It just goes one step further by making it a loop.
An algorithm using the same idea is shown in the accepted answer to your previous question:

Lines 1-8 generate a 1×1 square (by removing a corner) and then a 2×1 rectangle (by removing the corner), then a 3×1 rectangle, then a 4×1 rectangle, etc.
Line 9 stops the loop when the number of rectangles generated reaches a certain number $N$.
Line 10 subtracts 1 from $N$; this is the number of squares you have generated, or equivalently, the number of times you have looped through the loop of lines 1-8.

The difference is that the accepted answer generates the squares and rect