Interspace program in Haskell

Question

insert :: Eq(a) => a -> a -> [a] -> [a]
insert m n [] = []
insert m n (x:xs) | m==x = n : x : insert m n xs  
                 | otherwise = x : insert m n x         


        

Answer 1

Since you will only be adding values to the front of the list, your insert function is unnecessary. (:) will suffice instead. Much like in insert, we pass recursively over the list. Since we want to check if two values match at a time and will also call the function recursively on different lists based on whether or not we find a match, it's a good idea to pattern match (x1:x2:xs) rather than just (x:xs).

If m matches x1 and q matches x2, we place the onto the head of the list and call interspace recursively on the rest of the list. If they do not mach, we call interspace on (x2:xs).

 interspace :: Eq a => a -> a -> a-> [a] -> [a]
 interspace m n q []         = []
 interspace m n q [x]        = [x]
 interspace m n q (x1:x2:xs) | m == x1 && q == x2 = m : n : q : interspace m n q xs
                             | otherwise          = x1 : interspace m n q (x2:xs)

Example usage:

ghci>> interspace 1 2 1 [1,1,2,1,1]
[1,2,1,2,1,2,1]
ghci>> interspace 1 2 1 [1,1,3]
[1,2,1,3]
ghci>> interspace 1 2 1 [1,2,4,2]
[1,2,4,2]
ghci>> interspace 1 2 1 [1]
[1]
ghci>> interspace 1 2 1 []
[]