136 lines
4.1 KiB
Kotlin
136 lines
4.1 KiB
Kotlin
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import java.io.File
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typealias Hand = List<Char>
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fun main() {
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val input = File("input")
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val bets = parseHands(input)
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println(part1(bets))
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println(part2(bets))
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}
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fun part1(bets: List<Pair<Hand, Int>>): Int {
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return bets.sortedBy { handValue(it.first) }.foldIndexed(0) { i, acc, bet ->
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acc + (i + 1) * bet.second
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}
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}
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fun part2(bets: List<Pair<Hand, Int>>): Int {
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// println(
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// bets.sortedBy { handValueJoker(it.first) }.forEach {
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// println("${it} - ${handValueJoker(it.first)}")
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// }
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// )
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return bets.sortedBy { handValueJoker(it.first) }.foldIndexed(0) { i, acc, bet ->
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acc + (i + 1) * bet.second
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}
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}
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fun cardValue(card: Char): Int {
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return when (card) {
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'A' -> 12
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'K' -> 11
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'Q' -> 10
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'J' -> 9
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'T' -> 8
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in '2'..'9' -> card.digitToInt() - 2
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else -> throw Exception("Invalid card")
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}
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}
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fun type(cards: Hand): Int {
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return cards.groupBy { it }.mapValues { it.value.count() }.entries.fold(0) { acc, entry ->
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acc +
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when (entry.value) {
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5 -> 6000000
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4 -> 5000000
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3 -> 3000000
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2 -> 1000000
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1 -> 0
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else -> throw Exception("Too many cards")
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}
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}
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}
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fun handValue(cards: Hand): Int {
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// Each hand has 5 cards, the possible card values are 2-9+TJQKA,
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// and the order matters.
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//
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// This means the hands themselves are basically a 5-digit base 13
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// number.
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//
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// In addition, particular groupings of the cards matter. This is
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// represented by the type - to condense this into an easily
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// usable number, the type is therefore encoded in digits just
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// beyond the max of the 5-digit base 13 numbers. For convenient
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// programming, this means the type is encoded by the millions
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// digit in base 10, hence the odd type return numbers.
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return type(cards) + cards.fold(0) { acc, label -> acc * 13 + cardValue(label) }
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}
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fun cardValueJoker(card: Char): Int {
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return when (card) {
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'A' -> 12
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'K' -> 11
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'Q' -> 10
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'T' -> 9
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in '2'..'9' -> card.digitToInt() - 1
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'J' -> 0
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else -> throw Exception("Invalid card")
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}
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}
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fun typeJoker(cards: Hand): Int {
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var groups: MutableMap<Char, Int> =
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cards.groupBy { it }.mapValues { it.value.count() }.toMutableMap()
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// If there is only one group, we return early with the highest
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// value (this is to catch edge cases where there are only jokers)
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if (groups.entries.size == 1) {
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return 6000000
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}
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val jokers = groups.remove('J') ?: 0
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return groups.entries.sortedByDescending { 13 * it.value + cardValueJoker(it.key) }.foldIndexed(
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0
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) { i, acc, entry ->
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acc +
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when (entry.value + if (i == 0) jokers else 0) {
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5 -> 6000000
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4 -> 5000000
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3 -> 3000000
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2 -> 1000000
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1 -> 0
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else -> throw Exception("Too many cards")
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}
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}
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}
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fun handValueJoker(cards: Hand): Int {
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// This is just a bit more complicated; the card value doesn't
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// change much, but the *type* changes quite a bit.
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//
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// Rather than just giving the best type the highest value, we now
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// need to take into account scenarios in which ties are broken by
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// jokers. To do this, we sort the groups by the value of their
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// cards and the number of them.
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//
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// Since the tie breaker in this case is again the value of the
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// card, this is kind of recursive.
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return typeJoker(cards) + cards.fold(0) { acc, label -> acc * 13 + cardValueJoker(label) }
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// 247838255 is too low
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}
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fun parseHands(input: File): List<Pair<Hand, Int>> {
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return input.useLines {
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it
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.map {
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val (hand, bid) = it.split(' ')
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hand.toList() to bid.toInt()
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}
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.toList()
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}
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}
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