day4: problem 2 solution
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@ -77,6 +77,13 @@ impl Card {
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0
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}
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}
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// Return the number of copies won
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fn win_copies(&self) -> usize {
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self.winning_numbers
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.iter()
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.filter(|win| self.our_numbers.contains(win))
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.count()
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}
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}
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impl From<&str> for Card {
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@ -127,11 +134,87 @@ fn problem1(input: InputIter) -> u64 {
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.map(|line| Card::from(line.unwrap().as_str()))
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.collect();
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cards.iter().map(|x| x.score()).sum()
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}
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// PROBLEM 2 solution
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fn problem2(input: InputIter) -> u64 {
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0
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// --- Part Two ---
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// Just as you're about to report your findings to the Elf, one of you realizes that the
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// rules have actually been printed on the back of every card this whole time.
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// There's no such thing as "points". Instead, scratchcards only cause you to win more
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// scratchcards equal to the number of winning numbers you have.
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// Specifically, you win copies of the scratchcards below the winning card equal to the
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// number of matches. So, if card 10 were to have 5 matching numbers, you would win one
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// copy each of cards 11, 12, 13, 14, and 15.
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// Copies of scratchcards are scored like normal scratchcards and have the same card
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// number as the card they copied. So, if you win a copy of card 10 and it has 5
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// matching numbers, it would then win a copy of the same cards that the original card
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// 10 won: cards 11, 12, 13, 14, and 15. This process repeats until none of the copies
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// cause you to win any more cards. (Cards will never make you copy a card past the end
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// of the table.)
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// This time, the above example goes differently:
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// 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
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// 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 |
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// 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
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// 18 13 56 72 | 74 77 10 23 35 67 36 11
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// Card 1 has four matching numbers, so you win one copy each of the next four
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// cards: cards 2, 3, 4, and 5. Your original card 2 has two matching numbers, so
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// you win one copy each of cards 3 and 4. Your copy of card 2 also wins one copy
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// each of cards 3 and 4. Your four instances of card 3 (one original and three
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// copies) have two matching numbers, so you win four copies each of cards 4 and 5.
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// Your eight instances of card 4 (one original and seven copies) have one matching
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// number, so you win eight copies of card 5. Your fourteen instances of card 5 (one
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// original and thirteen copies) have no matching numbers and win no more cards.
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// Your one instance of card 6 (one original) has no matching numbers and wins no
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// more cards.
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// Once all of the originals and copies have been processed, you end up with 1 instance
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// of card 1, 2 instances of card 2, 4 instances of card 3, 8 instances of card 4, 14
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// instances of card 5, and 1 instance of card 6. In total, this example pile of
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// scratchcards causes you to ultimately have 30 scratchcards!
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// Process all of the original and copied scratchcards until no more scratchcards are
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// won. Including the original set of scratchcards, how many total scratchcards do you
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// end up with?
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fn problem2_award_copies<'a>(orig_cards: &'a Vec<Card>, card: &Card) -> Vec<&'a Card> {
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let win_copies = card.win_copies();
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let mut new_cards = Vec::new();
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if win_copies != 0 {
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for ofs in 0..win_copies {
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// card.id is 1-indexed, so would need to be card.id-1, but offset needs to
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// start at 1, which offsets this
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new_cards.push(&orig_cards[card.id + ofs]);
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}
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}
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new_cards
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}
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fn problem2(input: InputIter) -> u64 {
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let orig_cards: Vec<Card> = input
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.map(|line| Card::from(line.unwrap().as_str()))
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.collect();
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let mut new_cards = Vec::new();
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let mut won_cards = Vec::new();
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for card in &orig_cards {
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new_cards.append(&mut problem2_award_copies(&orig_cards, card))
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}
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while let Some(card) = new_cards.pop() {
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won_cards.push(card);
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new_cards.append(&mut problem2_award_copies(&orig_cards, card))
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}
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(won_cards.len() + orig_cards.len()) as u64
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}
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