I'm trying to write a function of the form
f :: String -> [String]
f str = ...
that returns the list of all the strings formed by removing exactly one character from str. For example:
ghci> f "stack"
["tack","sack","stck","stak","stac"]
Because String and [Char] are synonymous, I could use the index, but I know that you should avoid doing that in Haskell. Is there a better way besides using the index?
You could use recursion like so:
f :: [a] -> [[a]]
f [] = []
f (s:ss) = ss : map (s:) (f ss)
[] :: [b] and in this case b ~ [a].
:)f [] = [[]]. Then f "a" = f ('a':[]) = [] : map ('a':) (f []) = [] : map ('a':) ([[]]) = [] : ['a':[]] = [] : ["a"] = [ [], "a" ]The Josh Kirklin's solution as a one-liner:
f = tail . foldr (\x ~(r:rs) -> (x : r) : r : map (x :) rs) [[]]
f = snd . foldr (\x ~(r, rs) -> (x : r, r : map (x :) rs)) ([], [])Maybe a more readable way to describe it is:
gaps :: [a] -> [[a]]
gaps xs = zipWith removeAt [0..] $ replicate (length xs) xs
removeAt i xs = ys ++ zs
where
(ys,_:zs) = splitAt i xs
But practically, it is slower than the other solutions.
