πββοΈ - 2024 DAY 20 SOLUTIONS -πββοΈ
Day 20: Race Condition
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First parse and floodfill from start, each position then holds the distance from the start
For part 1, I check all neighbor tiles of neighbor tiles that are walls and calculate the distance that would've been in-between.
In part 2 I check all tiles within a manhattan distance <= 20 and calculate the distance in-between on the path.
Then filter out all cheats <100 and count
Takes 1.4s sadly, I believe there is still potential for optimization.
Edit: coding style
import Control.Arrow
import qualified Data.List as List
import qualified Data.Set as Set
import qualified Data.Map as Map
import qualified Data.Maybe as Maybe
parse s = Map.fromList [ ((y, x), c) | (l, y) <- zip ls [0..], (c, x) <- zip l [0..]]
where
ls = lines s
floodFill m = floodFill' m startPosition (Map.singleton startPosition 0)
where
startPosition = Map.assocs
>>> filter ((== 'S') . snd)
>>> head
>>> fst
$ m
neighbors (p1, p2) = [(p1-1, p2), (p1, p2-1), (p1, p2+1), (p1+1, p2)]
floodFill' m p f
| m Map.! p == 'E' = f
| otherwise = floodFill' m n f'
where
seconds = f Map.! p
ns = neighbors p
n = List.filter ((m Map.!) >>> (`Set.member` (Set.fromList ".E")))
>>> List.filter ((f Map.!?) >>> Maybe.isNothing)
>>> head
$ ns
f' = Map.insert n (succ seconds) f
taxiCabDistance (a1, a2) (b1, b2) = abs (a1 - b1) + abs (a2 - b2)
calculateCheatAdvantage f (p1, p2) = c2 - c1 - taxiCabDistance p1 p2
where
c1 = f Map.! p1
c2 = f Map.! p2
cheatDeltas :: Int -> Int -> [(Int, Int)]
cheatDeltas l h = [(y, x) | x <- [-h..h], y <- [-h..h], let d = abs x + abs y, d <= h, d >= l]
(a1, a2) .+. (b1, b2) = (a1 + b1, a2 + b2)
solve l h (f, ps) = Set.toList
>>> List.map ( repeat
>>> zip (cheatDeltas l h)
>>> List.map (snd &&& uncurry (.+.))
>>> List.filter (snd >>> (`Set.member` ps))
>>> List.map (calculateCheatAdvantage f)
>>> List.filter (>= 100)
>>> List.length
)
>>> List.sum
$ ps
part1 = solve 2 2
part2 = solve 1 20
main = getContents
>>= print
. (part1 &&& part2)
. (id &&& Map.keysSet)
. floodFill
. parse
Gotcha, thanks. I just re-read the problem statement and looked at the input and my input has the strongest possible version of that constraint: the path is unbranching and has start and end at the extremes. Thank-you!
(I ask because everyone's solution seems to make the same assumption - that is, that you're finding a shortcut onto the same path, as opposed to onto a different path.)
Some others have answered already, but yes, there was a well-hidden line in the problem description about the map having only a single path from start to end..