Scratchcards

Part 1

Looks like we have to count how many numbers are shared between two halves of a scratchcard. We should first figure out how to do this.

It’s a simple matter to get both halves of the scratchcard — no regex needed this time. And to count the matches, we can convert both scratchcard halves to sets, use the & operator (set intersection) to get the numbers in common, and get the len of the result.

def num_matching_numbers(line: str) -> int:
    _, rest = line.split(":")
    winners, nums = rest.split("|")
    # Count only the numbers in common between both lists
    return len(set(winners.split()) & set(nums.split()))

Once we know how many matches we have, we must use that to calculate the points we’ve earned. 0 matches earns us 0 points; otherwise, the first match earns us 1 point, and every match after that doubles our points. This pattern exactly matches the powers of 2; we just need to subtract 1 from the exponent, and use int() to have 0 matches result in 0 points instead of 0.5 points.

This leads to a fairly simple comprehension inside of a sum.

...

class Solution(StrSplitSolution):
    def part_1(self) -> int:
        return sum(
            int(2 ** (num_matching_numbers(line) - 1))
            for line in self.input
        )

Part 2

Apparently winning on scratchcards doesn’t earn you points, but only more scratchcards? Sounds pointless, but let’s figure this out anyway.

First, let’s use a defaultdict(int) to keep track of how many copies of each card we have. At the end, we’ll count our scratchcards by summing up the values we kept track of.

from collections import defaultdict

...

class Solution(StrSplitSolution):
    ...

    def part_2(self) -> int:
        num_cards: dict[int, int] = defaultdict(int)
        for card_id, line in enumerate(self.input, start=1):
            num_matches = num_matching_numbers(line)

            ...  # TODO Add win-more-scratchcards code

        return sum(num_cards.values())

There are only two ways to earn scratchcards:

1. By starting out with that scratchcard. (Each initial scratchcard counts as one scratchcard!)
2. By winning it from a match on a previous card.

Let’s say you’ve found $m$ matching numbers on scratchcard number $c$; the IDs of the cards you’ve won will be the next $m$ cards in line (from $c + 1$ to $c + m$), and the number of copies you get of each of these cards will be however many copies you have of card $c$. (The wording of the prompt is a bit confusing, but that’s as simple as I could describe it!)

...

class Solution(StrSplitSolution):
    ...

    def part_2(self) -> int:
        num_cards: dict[int, int] = defaultdict(int)
        for card_id, line in enumerate(self.input, start=1):
            num_matches = num_matching_numbers(line)

            # You start with 1 of this scratchcard
            num_cards[card_id] += 1
            # For each copy of this scratchcard, a copy of each of the
            # next `num_matches` scratchcards is won
            for c in range(card_id + 1, card_id + num_matches + 1):
                num_cards[c] += num_cards[card_id]

        return sum(num_cards.values())

Once the rules are clarified to that level of precision, implementing them is straightforward, and the resulting code is pretty terse!