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Easy haskell solution using the Data.Sequence interface. It uses the fact that the next row in pascal's triangle can be computed as: zipWith (+) (0:previous) (previous ++ [0]). Since accesing last element is not optimal solution for lists, Sequence interface fits better I think.

{-# language OverloadedLists #-}
import Data.Sequence (Seq (..), (<|), (|>))
import qualified Data.Sequence as S
import Data.Foldable (toList)
pascal :: Int -> Seq (Seq Int)
pascal n = S.iterateN n pascal' [1]
where pascal' s = S.zipWith (+) (0 <| s) (s |> 0)
main :: IO ()
main = do
n <- readLn
let pasc = toList $ pascal n
putStrLn `$ unlines $` map (unwords . map show . toList) $ pasc

## Pascal's Triangle

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Easy haskell solution using the

`Data.Sequence`

interface. It uses the fact that the next row in pascal's triangle can be computed as:`zipWith (+) (0:previous) (previous ++ [0])`

. Since accesing last element is not optimal solution for lists,`Sequence`

interface fits better I think.