Day 12: Hot Springs
Megathread guidelines
- Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
- Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ , pastebin, or github (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)
FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
🔒 Thread is locked until there’s at least 100 2 star entries on the global leaderboard
🔓 Unlocked after 25 mins
Scala3
def countDyn(a: List[Char], b: List[Int]): Long = // Simple dynamic programming approach // We fill a table T, where // T[ ai, bi ] -> number of ways to place b[bi..] in a[ai..] // T[ ai, bi ] = 0 if an-ai >= b[bi..].sum + bn-bi // T[ ai, bi ] = 1 if bi == b.size - 1 && ai == a.size - b[bi] - 1 // T[ ai, bi ] = // (place) T [ ai + b[bi], bi + 1] if ? or # // (skip) T [ ai + 1, bi ] if ? or . // def t(ai: Int, bi: Int, tbl: Map[(Int, Int), Long]): Long = if ai >= a.size then if bi >= b.size then 1L else 0L else val place = Option.when( bi < b.size && // need to have piece left ai + b(bi) <= a.size && // piece needs to fit a.slice(ai, ai + b(bi)).forall(_ != '.') && // must be able to put piece there (ai + b(bi) == a.size || a(ai + b(bi)) != '#') // piece needs to actually end )((ai + b(bi) + 1, bi + 1)).flatMap(tbl.get).getOrElse(0L) val skip = Option.when(a(ai) != '#')((ai + 1, bi)).flatMap(tbl.get).getOrElse(0L) place + skip @tailrec def go(ai: Int, tbl: Map[(Int, Int), Long]): Long = if ai == 0 then t(ai, 0, tbl) else go(ai - 1, tbl ++ b.indices.inclusive.map(bi => (ai, bi) -> t(ai, bi, tbl)).toMap) go(a.indices.inclusive.last + 1, Map()) def countLinePossibilities(repeat: Int)(a: String): Long = a match case s"$pattern $counts" => val p2 = List.fill(repeat)(pattern).mkString("?") val c2 = List.fill(repeat)(counts).mkString(",") countDyn(p2.toList, c2.split(",").map(_.toInt).toList) case _ => 0L def task1(a: List[String]): Long = a.map(countLinePossibilities(1)).sum def task2(a: List[String]): Long = a.map(countLinePossibilities(5)).sum
(Edit: fixed mangling of &<)
I’m struggling to fully understand your solution. Could you tell me, why do you return
1
when at the end ofa
andb
? And why do you start fromsize + 1
?T counts the number of ways to place the blocks with lengths specified in b in the remaining a.size - ai slots. If there are no more slots left, there are two cases: Either there are also no more blocks left, then everything is fine, and the current situation is 1 way to place the blocks in the slots. Otherwise, there are still blocks left, and no more space to place them in. This means the current sitution is incorrect, so we contribute 0 ways to place the blocks. This is what the
if bi >= b.size then 1L else 0L
does.The start at
size + 1
is necessary, as we need to compute every table entry before it may get looked up. When placing the last block, we may check the entry(ai + b(bi) + 1, bi + 1)
, whereai + b(bi)
may already equala.size
(in the case where the block ends exactly at the end ofa
). The+ 1
in the entry is necessary, as we need to skip a slot after every block: If we looked at(ai + b(bi), bi + 1)
, we could start ata.size
, but then, for e.g.b = [2, 3]
, we would consider...#####.
a valid placement.Let me know if there are still things unclear :)
Thanks for the detailed explanation. It helped a lot, especially what the
tbl
actually holds.I’ve read your code again and I get how it works, but it still feels kinda strange that we are considering values outside of range of
a
andb
, and that we are marking them as correct. Like in first row of the example???.### 1,1,3
, there is no spring at8
and no group at3
but we are marking(8,3)
and(7,3)
as correct. In my mind, first position that should be marked as correct is4,2
, because that’s where group of 3 can fit.If you make the recurrent case a little more complicated, you can sidestep the weird base cases, but I like reducing the endpoints down to things like this that are easily implementable, even if they sound a little weird at first.