Day 2: Gift Shop

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FAQ

  • VegOwOtenks@lemmy.world
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    3 days ago

    Easy one to get through, no edge-cases biting me this time.

    I learned this year again: running in interpreted mode can cause significant slowdowns. Later, I’ll hopefully find the time clean it up, this solution feels ugly. Reading everyone else did it also like this or with regex makes me feel better about it though.

    Haskell

    Code from this morning (AoC is at 06:00 at my place)
    module Main (main) where
    import qualified Text.ParserCombinators.ReadP as ReadP
    import Numeric.Natural (Natural)
    import Control.Monad ((<$!>), guard)
    import qualified Data.List as List
    import Control.Arrow ((>>>))
    import qualified Data.Text as Text
    import qualified Data.Foldable as Foldable
    
    newtype Range = Range { getRange :: (Natural, Natural) }
      deriving Show
    
    parseRange :: ReadP.ReadP Range
    parseRange = do
      n1 <- ReadP.readS_to_P reads
      _ <- ReadP.char '-'
      n2 <- ReadP.readS_to_P reads
      pure . Range $ (n1, n2)
    
    parseLine :: ReadP.ReadP [Range]
    parseLine = parseRange `ReadP.sepBy` ReadP.char ','
    
    main :: IO ()
    main = do
      ranges <- fst . last . ReadP.readP_to_S parseLine <$!> getContents
      print $ part1 ranges
      print $ part2 ranges
    
    part1 :: [Range] -> Natural
    part1 = List.concatMap (uncurry enumFromTo . getRange)
      >>> List.filter isDoublePattern
      >>> Foldable.sum
    
    part2 :: [Range] -> Natural
    part2 = List.concatMap (uncurry enumFromTo . getRange)
      >>> List.filter isMultiplePattern
      >>> Foldable.sum
    
    isMultiplePattern :: Natural -> Bool
    isMultiplePattern n = let
        textN = Text.show n
        textLength = Text.length textN
      in flip any (divisorsOf textLength) $ \ divisor -> let
          patternLength = textLength `div` divisor
          patternPart = Text.take (fromIntegral patternLength) textN
        in Text.replicate (fromIntegral divisor) patternPart == textN
    
    isDoublePattern :: Natural -> Bool
    isDoublePattern n = let
      textN = Text.show n
      evenLength = even (Text.length textN)
      (first, second) = Text.splitAt (Text.length textN `div` 2) textN
      in evenLength && first == second
    
    divisorsOf :: Integral b => b -> [b]
    divisorsOf n = do
      x <- [2..n]
      guard ((n `mod` x) == 0)
      pure x
    

    Using the interpreter, this solution made me wait for two minutes until I could submit. x.x After testing it again in compiled mode, it only takes four seconds.