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}")