advent-of-code/2021/day8.py
Mikaël Capelle f5aabbee8f 2021 day 8.
2023-12-10 20:23:22 +01:00

88 lines
2.1 KiB
Python

import itertools
import os
import sys
VERBOSE = os.getenv("AOC_VERBOSE") == "True"
digits = {
"abcefg": 0,
"cf": 1,
"acdeg": 2,
"acdfg": 3,
"bcdf": 4,
"abdfg": 5,
"abdefg": 6,
"acf": 7,
"abcdefg": 8,
"abcdfg": 9,
}
lines = sys.stdin.read().splitlines()
# part 1
lengths = {len(k) for k, v in digits.items() if v in (1, 4, 7, 8)}
answer_1 = sum(
len(p) in lengths for line in lines for p in line.split("|")[1].strip().split()
)
print(f"answer 1 is {answer_1}")
# part 2
values: list[int] = []
for line in lines:
parts = line.split("|")
broken_digits = sorted(parts[0].strip().split(), key=len)
per_length = {
k: list(v)
for k, v in itertools.groupby(sorted(broken_digits, key=len), key=len)
}
# a can be found immediately
a = next(u for u in per_length[3][0] if u not in per_length[2][0])
# c and f have only two possible values corresponding to the single entry of
# length 2
cf = list(per_length[2][0])
# the only digit of length 4 contains bcdf, so we can deduce bd by removing cf
bd = [u for u in per_length[4][0] if u not in cf]
# the 3 digits of length 5 have a, d and g in common
adg = [u for u in per_length[5][0] if all(u in pe for pe in per_length[5][1:])]
# we can remove a
dg = [u for u in adg if u != a]
# we can deduce d and g
d = next(u for u in dg if u in bd)
g = next(u for u in dg if u != d)
# then b
b = next(u for u in bd if u != d)
# f is in the three 6-length digits, while c is only in 2
f = next(u for u in cf if all(u in p for p in per_length[6]))
# c is not f
c = next(u for u in cf if u != f)
# e is the last one
e = next(u for u in "abcdefg" if u not in {a, b, c, d, f, g})
mapping = dict(zip((a, b, c, d, e, f, g), "abcdefg"))
value = 0
for number in parts[1].strip().split():
digit = "".join(sorted(mapping[c] for c in number))
value = 10 * value + digits[digit]
if VERBOSE:
print(value)
values.append(value)
answer_2 = sum(values)
print(f"answer 2 is {answer_2}")