# [Haskell-cafe] Fastest way to calculate all the ways to interleave two lists

David Feuer david.feuer at gmail.com
Sun Apr 3 03:05:19 UTC 2016

I ran into a fun question today: http://stackoverflow.com/q/36342967/1477667

Specifically, it asks how to find all ways to interleave lists so that
the order of elements within each list is preserved. The most
efficient way I found is copied below. It's nicely lazy, and avoids
left-nested appends. Unfortunately, it's pretty seriously ugly. Does
anyone have any idea of a way to do this that's both efficient and
elegant?

{-# LANGUAGE BangPatterns #-}
import Data.Monoid
import Data.Foldable (toList)
import Data.Sequence (Seq, (|>))

-- Find all ways to interleave two lists
interleave2 :: [a] -> [a] -> [[a]]
interleave2 xs ys = interleave2' mempty xs ys []

-- Find all ways to interleave two lists, adding the
-- given prefix to each result and continuing with
-- a given list to append
interleave2' :: Seq a -> [a] -> [a] -> [[a]] -> [[a]]
interleave2' !prefix xs ys rest =
(toList prefix ++ xs ++ ys)
: interleave2'' prefix xs ys rest

-- Find all ways to interleave two lists except for
-- the trivial case of just appending them. Glom
-- the results onto the given list.
interleave2'' :: Seq a -> [a] -> [a] -> [[a]] -> [[a]]
interleave2'' !prefix [] _ = id
interleave2'' !prefix _ [] = id
interleave2'' !prefix xs@(x : xs') ys@(y : ys') =
interleave2' (prefix |> y) xs ys' .
interleave2'' (prefix |> x) xs' ys

-- What the question poser wanted; I don't *think* there's
-- anything terribly interesting to do here.
interleavings :: [[a]] -> [[a]]
interleavings = foldr (concatMap . interleave2) [[]]

Thanks,
David