Advent of Code 2023 - Day 4

Dec 4, 2023 Serie: AoC 2023

You can find the final code @ advent-of-code/2023/day4

In Day 4 we have the following input

Card 1: 41 48 83 86 17 | 83 86  6 31 17  9 48 53
Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19
Card 3:  1 21 53 59 44 | 69 82 63 72 16 21 14  1
Card 4: 41 92 73 84 69 | 59 84 76 51 58  5 54 83
Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36
Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11

In the above example, card 1 has five winning numbers (41, 48, 83, 86, and 17) and eight numbers you have (83, 86, 6, 31, 17, 9, 48, and 53). Of the numbers you have, four of them (48, 83, 17, and 86) are winning numbers! That means card 1 is worth 8 points (1 for the first match, then doubled three times for each of the three matches after the first).

The parsing logic is very similar to day 2, let’s create a struct that contains the winning numbers and the contained numbers of a card. This time I’m using a HashSet since it’s faster to check if a value is contained into another set.

#[derive(Debug)]
struct Card {
    winning: HashSet<u32>,
    numbers: HashSet<u32>,
}

impl From<&str> for Card {
    fn from(value: &str) -> Self {
        let (_, all_nums) = value.split_once(":").unwrap();
        let (winning_str, numbers_str) = all_nums.split_once("|").unwrap();

        let winning: HashSet<u32> = winning_str
            .trim()
            .replace("  ", " ")
            .split(' ')
            .map(|x| x.parse().unwrap())
            .collect();

        let numbers: HashSet<u32> = numbers_str
            .trim()
            .replace("  ", " ")
            .split(' ')
            .map(|x| x.parse().unwrap())
            .collect();

        Card { winning, numbers }
    }
}

I want to create a function that counts the matching numbers for a certain card

impl Card {
    fn matching(&self) -> u32 {
        let mut matching = 0;
        for w in &self.winning {
            if self.numbers.contains(w) {
                matching += 1;
            }
        }

        matching
    }
}

After that I need a function to actually calculate the points of each card, which depends on the number of matching winning numbers. The first match gives the card 1 point, after that we have to multiply the card’s points by 2, which is equivalent to pow the card points by 2. Looks like a good case where bit shifting can be used

impl Card {
    fn points(&self) -> u32 {
        let matching = self.matching();
        if matching == 0 {
            return 0;
        }
        1 << (matching - 1)
    }
}

That is all we need for part1, let’s run some tests

fn part1(input: &str) -> io::Result<u32> {
    let points = input.lines().map(Card::from).map(|x| x.points()).sum();

    Ok(points)
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_part1() {
        assert_eq!(
            13,
            part1(
                "Card 1: 41 48 83 86 17 | 83 86  6 31 17  9 48 53
Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19
Card 3:  1 21 53 59 44 | 69 82 63 72 16 21 14  1
Card 4: 41 92 73 84 69 | 59 84 76 51 58  5 54 83
Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36
Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11"
            )
            .unwrap()
        );
    }
}

Aaand part1 solution is correct! Let’s move on with part 2

The problem states

There’s no such thing as "points". Instead, scratchcards only cause you to win more scratchcards equal to the number of winning numbers you have.

Specifically, you win copies of the scratchcards below the winning card equal to the number of matches. So, if card 10 were to have 5 matching numbers, you would win one copy each of cards 11, 12, 13, 14, and 15.

Copies of scratchcards are scored like normal scratchcards and have the same card number as the card they copied. So, if you win a copy of card 10 and it has 5 matching numbers, it would then win a copy of the same cards that the original card 10 won: cards 11, 12, 13, 14, and 15. This process repeats until none of the copies cause you to win any more cards. (Cards will never make you copy a card past the end of the table.)

We can reuse at least the matching function of Card. We now need to calculate how may cards we end up.

fn part2(input: &str) -> io::Result<u32> {
    let matches: Vec<u32> = input
        .lines()
        .map(Card::from)
        .map(|x| x.matching())
        .collect();

    let mut cards: Vec<u32> = vec![1; matches.len()];

    for (i, matching) in matches.iter().enumerate() {
        let index = i as u32 + 1;
        let incr = cards[i];

        // increment the number of cards that are in the window
        // (i+1) until ((i+1) + matching) by the number of cards
        // at position i
        for l in index..index + matching {
            if let Some(v) = cards.get_mut(l as usize) {
                *v += incr;
            }
        }
    }

    Ok(cards.into_iter().sum())
}

I am using a vector that keeps track of how many cards I have in my deck, i.e if cards[10] = 5 I have 5 Card 11, the index mismatch is not important in this case since we need to sum all the number of cards that we have in the deck at the end.

Note that I am using cards.get_mut because I can only add cards that are actually present in the deck. So, if the last card in the deck is Card 6 and Card 6 has 2 matching values I cannot add Card 7 and Card 8, because they’re not present in the deck, in that case cards.get_mut will return None and no operation is going to be performed.

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_part2() {
        assert_eq!(
            30,
            part2(
                "Card 1: 41 48 83 86 17 | 83 86  6 31 17  9 48 53
Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19
Card 3:  1 21 53 59 44 | 69 82 63 72 16 21 14  1
Card 4: 41 92 73 84 69 | 59 84 76 51 58  5 54 83
Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36
Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11"
            )
            .unwrap()
        );
    }
}

Test passes and the solution is correct!