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And we're off!
I wonder if I got a bit too clever by half in this solution, as I've been looking over older solutions and can't even remember solving them. But that's how it goes I guess.
Puzzle rating: 3/5
Score: 2
A "standard" Perlish solution (well, *my* kind of Perl, anyway): a dispatch table for the else/if "switch" construct, and a compact hash containing the state of the two solutions.
Puzzle rating: 3/5
Score: 2
I was honestly surprised that the canonical solution to this wasn't some esoteric bit-twiddling trick that reduces it to a one-liner.
In part 2, the naive solution is to loop through each "column" to determine which values to count so as to determine whether they are most frequent or not. I used an index for each "set" to keep track of the values already assigned to that set.
Puzzle rating: 3/5
Score: 2
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Updated on Friday, 2021-12-03
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