# 99 questions/Solutions/17

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< 99 questions | Solutions(Difference between revisions)

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<haskell> | <haskell> | ||

split :: [a] -> Int -> ([a], [a]) | split :: [a] -> Int -> ([a], [a]) | ||

− | split xs 0 = ([], xs) | + | split (x:xs) n | n > 0 = let (f,l) = split xs (n-1) in (x : f, l) |

− | split (x:xs) n = | + | split xs _ = ([], xs) |

+ | </haskell> | ||

+ | |||

+ | Or (ab)using the "&&&" arrow operator for tuples: | ||

+ | |||

+ | <haskell> | ||

+ | split :: [a] -> Int -> ([a], [a]) | ||

+ | split (x:xs) n | n > 0 = (:) x . fst &&& snd $ split xs (n - 1) | ||

+ | split xs _ = ([], xs) | ||

+ | </haskell> | ||

+ | |||

+ | A similar solution using foldl: | ||

+ | |||

+ | <haskell> | ||

+ | split :: [a] -> Int -> ([a], [a]) | ||

+ | split [] _ = ([], []) | ||

+ | split list n | ||

+ | | n < 0 = (list, []) | ||

+ | | otherwise = (first output, second output) | ||

+ | where output = foldl (\acc e -> if third acc > 0 then (first acc ++ [e], second acc, third acc - 1) else (first acc, second acc ++ [e], third acc)) ([], [], n) list | ||

+ | </haskell> | ||

+ | |||

+ | Note that for the above code to work you must define your own first, second, and third functions for tuples containing three elements like so: | ||

+ | |||

+ | <haskell> | ||

+ | first :: (a, b, c) -> a | ||

+ | first (x, _, _) = x | ||

+ | |||

+ | second :: (a, b, c) -> b | ||

+ | second (_, y, _) = y | ||

+ | |||

+ | third :: (a, b, c) -> c | ||

+ | third (_, _, z) = z | ||

+ | </haskell> | ||

+ | |||

+ | Another foldl solution without defining tuple extractors: | ||

+ | <haskell> | ||

+ | split :: [a] -> Int -> ([a],[a]) | ||

+ | split lst n = snd $ foldl helper (0,([],[])) lst | ||

+ | where helper (i,(left,right)) x = if i >= n then (i+1,(left,right++[x])) else (i+1,(left++[x],right)) | ||

</haskell> | </haskell> |

## Revision as of 20:44, 8 April 2013

(*) Split a list into two parts; the length of the first part is given.

Do not use any predefined predicates.

Solution usingtake

drop

split xs n = (take n xs, drop n xs)

splitAt

split = flip splitAt

But these should clearly be considered "predefined predicates". Alternatively, we have the following recursive solution:

split :: [a] -> Int -> ([a], [a]) split [] _ = ([], []) split l@(x : xs) n | n > 0 = (x : ys, zs) | otherwise = ([], l) where (ys,zs) = split xs (n - 1)

The same solution as above written more cleanly:

split :: [a] -> Int -> ([a], [a]) split (x:xs) n | n > 0 = let (f,l) = split xs (n-1) in (x : f, l) split xs _ = ([], xs)

Or (ab)using the "&&&" arrow operator for tuples:

split :: [a] -> Int -> ([a], [a]) split (x:xs) n | n > 0 = (:) x . fst &&& snd $ split xs (n - 1) split xs _ = ([], xs)

A similar solution using foldl:

split :: [a] -> Int -> ([a], [a]) split [] _ = ([], []) split list n | n < 0 = (list, []) | otherwise = (first output, second output) where output = foldl (\acc e -> if third acc > 0 then (first acc ++ [e], second acc, third acc - 1) else (first acc, second acc ++ [e], third acc)) ([], [], n) list

Note that for the above code to work you must define your own first, second, and third functions for tuples containing three elements like so:

first :: (a, b, c) -> a first (x, _, _) = x second :: (a, b, c) -> b second (_, y, _) = y third :: (a, b, c) -> c third (_, _, z) = z

Another foldl solution without defining tuple extractors:

split :: [a] -> Int -> ([a],[a]) split lst n = snd $ foldl helper (0,([],[])) lst where helper (i,(left,right)) x = if i >= n then (i+1,(left,right++[x])) else (i+1,(left++[x],right))