I can't code today because of travel, just as the one that might run slow enough to geek out on benchmarking the various options. Does seem like it's a bit of a jump in difficulty from previous days.
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Based on the star distribution today (everyone who has solved part 1 also solved part2) I suspect at least part 2 is easy even if I can't make a dent on part1 yet
It was easy staight forward brtue force:
Spoiler
For each possible_rectange Find Max:
If polygon.contains(rectangle) Then rectangle.area() Else 0
For speed the loop can be run paralelley.
I had no time to implent the contains myself, but pretty much all languages have a library already.
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yea becomes trivial if one uses a polygon library (I did too) but looking at the actual polygon, one can cook up some clever heuristics to do it quite quickly with some basic checks I think.
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