After this, I approached this problem in a similar way as I did for problem 199.

I start with the point (1, 0) and I generate it's maximum width using the above, which is ~0.61, then I generate the next 2 rectangle which have bottom left point (1.61, 0) and (1, 0.61).

The next problem is finding out when to stop generating new squares.

I'm sure there is a smarter way, but I just set a limit to how small the width can be and made it smaller until I found the correct answer.

To implement all of this while keeping track of the index I implemented a stack (I also tried recursion but it was more difficult and slower). I outline the first case below

1. Elements in the stack look like this [a, b, left, under, min_x]

   • (a, b) is the bottom left point of a square

   • left, under denotes the number of squares to the left and under

   • min_x is the limit for width as I explained above

2. Start with [1, 0, 0, 0, min_x]

   • a = 1, b = 0, left = 0, under = 0

3. find the width, ~0.61, using above formula and given point

4. if the width is greater than min_x

   • add [1 + 0.61, 0, 1, 0, min_x] notice we increased under by 1 because this square will be above the first square

   • add [1, 0.61, 0, 1, min_x] notice we increased left by 1 because this square will be to the left of the first square

After all this just loop through a sorted list and find the desired answer