Day 3: Mull It Over

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FAQ

  • @[email protected]
    3•5 months ago

    J

    We can take advantage of the manageable size of the input to avoid explicit looping and mutable state; instead, construct vectors which give, for each character position in the input, the position of the most recent do() and most recent don't(); for part 2 a multiplication is enabled if the position of the most recent do() (counting start of input as 0) is greater than that of the most recent don't() (counting start of input as minus infinity).

    load 'regex'

raw =: fread '3.data'
mul_matches =: 'mul\(([[:digit:]]{1,3}),([[:digit:]]{1,3})\)' rxmatches raw

NB. a b sublist y gives elements [a..b) of y
sublist =: ({~(+i.)/)~"1 _

NB. ". is number parsing
mul_results =: */"1 ". (}."2 mul_matches) sublist raw
result1 =: +/ mul_results

do_matches =: 'do\(\)' rxmatches raw
dont_matches =: 'don''t\(\)' rxmatches raw
match_indices =: (<0 0) & {"2
do_indices =: 0 , match_indices do_matches  NB. start in do mode
dont_indices =: match_indices dont_matches
NB. take successive diffs, then append length from last index to end of string
run_lengths =: (}. - }:) , (((#raw) & -) @: {:)
do_map =: (run_lengths do_indices) # do_indices
dont_map =: (({. , run_lengths) dont_indices) # __ , dont_indices
enabled =: do_map > dont_map
result2 =: +/ ((match_indices mul_matches) { enabled) * mul_results