Mikaël Capelle
ce315b8778
Co-authored-by: Mikael CAPELLE <mikael.capelle@thalesaleniaspace.com> Co-authored-by: Mikaël Capelle <capelle.mikael@gmail.com> Reviewed-on: #3
90 lines
2.5 KiB
Python
90 lines
2.5 KiB
Python
import itertools
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from typing import Any, Iterator
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from ..base import BaseSolver
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digits = {
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"abcefg": 0,
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"cf": 1,
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"acdeg": 2,
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"acdfg": 3,
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"bcdf": 4,
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"abdfg": 5,
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"abdefg": 6,
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"acf": 7,
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"abcdefg": 8,
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"abcdfg": 9,
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}
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class Solver(BaseSolver):
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def solve(self, input: str) -> Iterator[Any]:
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lines = input.splitlines()
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# part 1
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lengths = {len(k) for k, v in digits.items() if v in (1, 4, 7, 8)}
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yield sum(
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len(p) in lengths
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for line in lines
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for p in line.split("|")[1].strip().split()
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)
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# part 2
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values: list[int] = []
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for line in lines:
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parts = line.split("|")
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broken_digits = sorted(parts[0].strip().split(), key=len)
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per_length = {
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k: list(v)
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for k, v in itertools.groupby(sorted(broken_digits, key=len), key=len)
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}
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# a can be found immediately
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a = next(u for u in per_length[3][0] if u not in per_length[2][0])
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# c and f have only two possible values corresponding to the single entry of
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# length 2
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cf = list(per_length[2][0])
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# the only digit of length 4 contains bcdf, so we can deduce bd by removing cf
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bd = [u for u in per_length[4][0] if u not in cf]
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# the 3 digits of length 5 have a, d and g in common
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adg = [
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u for u in per_length[5][0] if all(u in pe for pe in per_length[5][1:])
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]
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# we can remove a
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dg = [u for u in adg if u != a]
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# we can deduce d and g
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d = next(u for u in dg if u in bd)
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g = next(u for u in dg if u != d)
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# then b
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b = next(u for u in bd if u != d)
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# f is in the three 6-length digits, while c is only in 2
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f = next(u for u in cf if all(u in p for p in per_length[6]))
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# c is not f
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c = next(u for u in cf if u != f)
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# e is the last one
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e = next(u for u in "abcdefg" if u not in {a, b, c, d, f, g})
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mapping = dict(zip((a, b, c, d, e, f, g), "abcdefg"))
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value = 0
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for number in parts[1].strip().split():
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digit = "".join(sorted(mapping[c] for c in number))
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value = 10 * value + digits[digit]
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self.logger.info(f"value for '{line}' is {value}")
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values.append(value)
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yield sum(values)
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