# --- Day 4: Passport Processing ---

# You arrive at the airport only to realize that you grabbed your North Pole
# Credentials instead of your passport. While these documents are extremely
# similar, North Pole Credentials aren't issued by a country and therefore
# aren't actually valid documentation for travel in most of the world.

# It seems like you're not the only one having problems, though; a very long
# line has formed for the automatic passport scanners, and the delay could
# upset your travel itinerary.

# Due to some questionable network security, you realize you might be able to
# solve both of these problems at the same time.

# The automatic passport scanners are slow because they're having trouble
# detecting which passports have all required fields. The expected fields are
# as follows:

#     byr (Birth Year)
#     iyr (Issue Year)
#     eyr (Expiration Year)
#     hgt (Height)
#     hcl (Hair Color)
#     ecl (Eye Color)
#     pid (Passport ID)
#     cid (Country ID)

# Passport data is validated in batch files (your puzzle input). Each passport
# is represented as a sequence of key:value pairs separated by spaces or
# newlines. Passports are separated by blank lines.

# Here is an example batch file containing four passports:

# ecl:gry pid:860033327 eyr:2020 hcl:#fffffd
# byr:1937 iyr:2017 cid:147 hgt:183cm

# iyr:2013 ecl:amb cid:350 eyr:2023 pid:028048884
# hcl:#cfa07d byr:1929

# hcl:#ae17e1 iyr:2013
# eyr:2024
# ecl:brn pid:760753108 byr:1931
# hgt:179cm

# hcl:#cfa07d eyr:2025 pid:166559648
# iyr:2011 ecl:brn hgt:59in

# The first passport is valid - all eight fields are present. The second
# passport is invalid - it is missing hgt (the Height field).

# The third passport is interesting; the only missing field is cid, so it looks
# like data from North Pole Credentials, not a passport at all! Surely, nobody
# would mind if you made the system temporarily ignore missing cid fields.
# Treat this "passport" as valid.

# The fourth passport is missing two fields, cid and byr. Missing cid is fine,
# but missing any other field is not, so this passport is invalid.

# According to the above rules, your improved system would report 2 valid
# passports.

# Count the number of valid passports - those that have all required fields.
# Treat cid as optional. In your batch file, how many passports are valid?


with open("files/P4.txt", "r") as f:
    passports = [line.split() for line in f.read().strip().split("\n\n")]


def part_1() -> None:
    valid_passports = 0
    for passport in passports:
        if len(passport) > 7:
            valid_passports += 1
        if len(passport) == 7:
            valid_fields = 0
            for field in passport:
                if "cid" not in field:
                    valid_fields += 1
            if valid_fields == 7:
                valid_passports += 1

    print(f"There are {valid_passports} valid passports")


# --- Part Two ---

# The line is moving more quickly now, but you overhear airport security
# talking about how passports with invalid data are getting through. Better add
# some data validation, quick!

# You can continue to ignore the cid field, but each other field has strict
# rules about what values are valid for automatic validation:

#     byr (Birth Year) - four digits; at least 1920 and at most 2002.
#     iyr (Issue Year) - four digits; at least 2010 and at most 2020.
#     eyr (Expiration Year) - four digits; at least 2020 and at most 2030.
#     hgt (Height) - a number followed by either cm or in:
#         If cm, the number must be at least 150 and at most 193.
#         If in, the number must be at least 59 and at most 76.
#     hcl (Hair Color) - a # followed by exactly six characters 0-9 or a-f.
#     ecl (Eye Color) - exactly one of: amb blu brn gry grn hzl oth.
#     pid (Passport ID) - a nine-digit number, including leading zeroes.
#     cid (Country ID) - ignored, missing or not.

# Your job is to count the passports where all required fields are both present
# and valid according to the above rules. Here are some example values:

# byr valid:   2002
# byr invalid: 2003

# hgt valid:   60in
# hgt valid:   190cm
# hgt invalid: 190in
# hgt invalid: 190

# hcl valid:   #123abc
# hcl invalid: #123abz
# hcl invalid: 123abc

# ecl valid:   brn
# ecl invalid: wat

# pid valid:   000000001
# pid invalid: 0123456789

# Here are some invalid passports:

# eyr:1972 cid:100
# hcl:#18171d ecl:amb hgt:170 pid:186cm iyr:2018 byr:1926

# iyr:2019
# hcl:#602927 eyr:1967 hgt:170cm
# ecl:grn pid:012533040 byr:1946

# hcl:dab227 iyr:2012
# ecl:brn hgt:182cm pid:021572410 eyr:2020 byr:1992 cid:277

# hgt:59cm ecl:zzz
# eyr:2038 hcl:74454a iyr:2023
# pid:3556412378 byr:2007

# Here are some valid passports:

# pid:087499704 hgt:74in ecl:grn iyr:2012 eyr:2030 byr:1980
# hcl:#623a2f

# eyr:2029 ecl:blu cid:129 byr:1989
# iyr:2014 pid:896056539 hcl:#a97842 hgt:165cm

# hcl:#888785
# hgt:164cm byr:2001 iyr:2015 cid:88
# pid:545766238 ecl:hzl
# eyr:2022

# iyr:2010 hgt:158cm hcl:#b6652a ecl:blu byr:1944 eyr:2021 pid:093154719

# Count the number of valid passports - those that have all required fields and
# valid values. Continue to treat cid as optional. In your batch file, how many
# passports are valid?


def is_valid(passport: dict[str, str]) -> bool:
    keys = ["byr", "iyr", "eyr", "hgt", "hcl", "ecl", "pid"]
    for key in keys:
        if key not in passport:
            return False
    if not any([passport[key] != 4 for key in ["byr", "iyr", "eyr"]]):
        return False
    if passport["byr"] < "1920" or passport["byr"] > "2002":
        return False
    if passport["iyr"] < "2010" or passport["iyr"] > "2020":
        return False
    if passport["eyr"] < "2020" or passport["eyr"] > "2030":
        return False
    if not any(["cm" in passport["hgt"], "in" in passport["hgt"]]):
        return False
    _hgt = int(passport["hgt"][:-2])
    if "cm" in passport["hgt"]:
        if _hgt < 150 or _hgt > 193:
            return False
    if "in" in passport["hgt"]:
        if _hgt < 59 or _hgt > 76:
            return False
    if "#" not in passport["hcl"]:
        return False
    _hcl = passport["hcl"].split("#")[-1]
    if len(_hcl) != 6:
        return False
    try:
        int(_hcl, 16)
    except ValueError:
        return False
    if not any(
        [
            passport["ecl"]
            in [
                "amb",
                "blu",
                "brn",
                "gry",
                "grn",
                "hzl",
                "oth",
            ]
        ]
    ):
        return False
    if len(passport["pid"]) != 9:
        return False
    return True


def part_2() -> None:
    valid_passports = 0
    for passport in passports:
        _items: list[list[str]] = [
            item.split(":") for item in passport if item
        ]
        passport_dict: dict[str, str] = dict(_items)
        if is_valid(passport_dict):
            valid_passports += 1
    print(f"There are {valid_passports} valid passports")


if __name__ == "__main__":
    part_1()
    part_2()