adventofcode2023/day07-2.py

from collections import Counter

from helpers import Helper

helper = Helper(debug=True)
debug = helper.debug
load_input = helper.load_input

CARD_VALUES = ["J"]
for i in range(2,10):
    CARD_VALUES.append(str(i))
CARD_VALUES.extend(["T", "Q", "K", "A"])

HAND_VALUES = [
    "High card",
    "One pair",
    "Two pair",
    "Three of a kind",
    "Full House",
    "Four of a kind",
    "Five of a kind"
]

class Hand:
    def __init__(self, cards, bid):
        self.cards = cards
        self.bid = bid

    @property
    def value(self):
        card_counter = Counter(self.cards)
        card_counts = card_counter.most_common()
        # We have to separate J being the most common card out,
        # or else it might get counted twice.
        if card_counts[0][0] == "J":
            # Five jacks is five of a kind.
            # Four jacks is five of a kind too, because the other card counts.
            # Three jacks plus two of another card is five of a kind.
            if (card_counts[0][1] == 5 
                or card_counts[0][1] == 4
                or (card_counts[0][1] == 3 and card_counts[1][1] == 2)
            ):
                return "Five of a kind"
            # Three jacks is four of a kind, because the next most common card counts.
            # Two jacks plus two of another card is four of a kind.
            if (card_counts[0][1] == 3
                or (card_counts[0][1] == 2 and card_counts[1][1] == 2)
            ):
                return "Four of a kind"
            # Weirdly, you can only get a full house if there's only one jack...
            # and if that's the case, it won't be at the front of most_common().
            if card_counts[0][1] == 2:
                return "Three of a kind"
            # If J is at the front of most_common(), and there's only one of it,
            # then there's no more than 1 of any other card. So we have a pair.
            return "One pair"
            
        # Okay, done with J being at the front.
        # There are five cards of a kind, or (three/four) cards of a kind and 
        # (one/two) J.
        # If there are only two of a kind at the front, then there aren't more
        # than two of any other kind, so we can't get to 5.
        if card_counts[0][1] == 5 or card_counts[0][1] + card_counter["J"] == 5:
            return "Five of a kind"
        # There are four cards of a kind without a J, or (two/three) cards of 
        # a kind and (one/two) J.
        if card_counts[0][1] == 4 or card_counts[0][1] + card_counter["J"] == 4:
            return "Four of a kind"
        # There are three cards of a kind without a J, or two cards of a kind and
        # exactly one J.
        if card_counts[0][1] == 3 or card_counts[0][1] + card_counter["J"] == 3:
            # We know the most common card isn't a J; we already covered that
            # in a separate branch. 
            # If the most common count is 3 and there's a J, we already covered
            # that too, with four and five of a kind.
            # If the most common count is 2 and there are 2 Js, we already covered
            # that with four of a kind.
            # So if the most common count is 2 and there's a J, it can't be in second
            # position, and if the most common count is 3 then there can't be a J.
            # So we can discard the possibility that J is in second position.
            if card_counts[1][1] == 2:
                return "Full House"
            return "Three of a kind"
        # There are two of the most common card without any Js, or one and a single J.
        if card_counts[0][1] == 2 or card_counts[0][1] + card_counter["J"] == 2:
            # Same logic as above. If J were the second-most-common card we'd have
            # already encountered it. 
            if card_counts[1][1] == 2:
                return "Two pair"
            return "One pair"
        return "High card"

    def __lt__(self, other):
        # They have different hand values
        if self.value != other.value:
            return HAND_VALUES.index(self.value) < HAND_VALUES.index(other.value)
        # They have the same hand value
        # So check each card in sequence
        for i in range(len(self.cards)):
            if self.cards[i] != other.cards[i]:
                # They have different nth cards
                return CARD_VALUES.index(self.cards[i]) < CARD_VALUES.index(other.cards[i])
        # They're the same
        return False
    
    def __str__(self):
        return f"Camel Cards hand: {self.cards} (value {self.value}, bid {self.bid})"
    
    def __repr__(self):
        return self.__str__()
        

def main():
    lines = load_input(7)
    hands = []
    for line in lines:
        cards, bid = line.split()
        hands.append(Hand(
            cards = cards,
            bid = bid
        ))
    hands.sort()
    print(hands[:10])

    total_winnings = 0
    for i, hand in enumerate(hands):
        total_winnings += ((i+1) * int(hand.bid))

    print(f"Total winnings: {total_winnings}")

if __name__ == "__main__":
    main()