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Filter Duplicate Elements in Haskell

Submitted by: @import:stackexchange-codereview··
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duplicatefilterelementshaskell

Problem

I'm working on HackerRank to try to improve my Haskell skills along side with reading Haskell Programming from first principles. I wrote a program that works, but it seems to time out on large input sets. The purpose of the program is

Given a list of n integers a = [a1, a2, ..., an], you have to find those integers which are repeated at least k times. In case no such element exists you have to print -1.

If there are multiple elements in a which are repeated at least k times, then print these elements ordered by their first occurrence in the list.

So I wrote a few different functions to help with this.
count which counts the number of occurrences of an element in a list
count :: Eq a => Integral b => a -> [a] -> b
count e [] = 0
count e (a:xs) = (count e xs +) $ if a == e then 1 else 0

uniq which removes duplicates from a list
uniq :: Eq a => [a] -> [a] -> [a]
uniq x [] = x
uniq [] (a:xs) = uniq [a] xs
uniq x (a:xs) = if a
elem x then uniq x xs else uniq (a:x) xs

filt which filters through a list and removes elements that don't occur at least k times.
filt :: Show a => Num a => Read a => Eq a => Integral b => [a] -> b -> [a]
filt a b = reverse $ uniq [] [i | i = b]

printList which prints a list as a space separated list or prints -1 if the list is empty.
printList :: Show a => [a] -> IO ()
printList [] = putStrLn "-1"
printList a = putStrLn $ unwords [show i | i
readNumbers which takes a space separated string and returns a Num list from that string.
readNumbers :: Show a => Eq a => Num a => Read a => String -> [a]
readNumbers = map read . words

run which throws all of this together and runs this n times.
run :: (Show a, Eq a, Num a, Read a) => a -> IO ()
run 0 = putStr ""
run n = do
a
main the main function. It gets a number n and then calls run n.
main :: IO ()
main = do
a

This code works, for example, with the input
3
9 2
4 5 2 5 4 3 1 3 4
9 4
4 5 2 5 4 3 1 3 4
10 2
5 4 3 2 1 1 2 3 4 5
`

and giv

Solution

If you write uniq as a right fold, you don't need to pass an accumulator through, and the list comes out in the right order:

uniq :: Eq a => [a] -> [a]
uniq [] = []
uniq (x:xs) = (if x `elem` xs then id else (x:)) $ uniq xs

filt :: Show a => Num a => Read a => Eq a => Integral b => [a] -> b -> [a]
filt k is = uniq [i | i = k]


(Edit: Actually that one throws out the first of each two equal elements, not the last. Heres one without that problem:

uniq :: Eq a => [a] -> [a]
uniq [] = []
uniq (x:xs) = x : uniq (filter (/=x) xs)


)

You've commendably already brought
count into a form that allows it to be written in terms of library combinators:

count :: Eq a => Integral b => a -> [a] -> b
count e = sum . map (\a -> if a == e then 1 else 0)


That's a bit ugly due to lambdas though, here's a nicer version:

count e = length . filter (== e)


For separation of monadic and pure code (and generally for factoring out common code from across cases), here's a
showList to replace printList:

showList :: Show a => [a] -> String
showList [] = "-1"
showList a = unwords [show i | i <- a]


Calling a monadic action a given number of times doesn't need manual recursion, and thus also doesn't need to give the repeated action a name:

main :: IO () 
main = do
  a  getLine
    numbers  getLine
    putStrLn $ showList $ filt k numbers


(I think
readNumbers doesn't deserve a name.)

In case the order in which the output is given isn't important, here's a version that doesn't require quadratic time because each element is compared to every other:

filt k = map head . filter ((>=k) . length) . group . sort


which relies on
Data.Lists sort` being faster than quadratic time.

Code Snippets

uniq :: Eq a => [a] -> [a]
uniq [] = []
uniq (x:xs) = (if x `elem` xs then id else (x:)) $ uniq xs

filt :: Show a => Num a => Read a => Eq a => Integral b => [a] -> b -> [a]
filt k is = uniq [i | i <- is, count i is >= k]
uniq :: Eq a => [a] -> [a]
uniq [] = []
uniq (x:xs) = x : uniq (filter (/=x) xs)
count :: Eq a => Integral b => a -> [a] -> b
count e = sum . map (\a -> if a == e then 1 else 0)
count e = length . filter (== e)
showList :: Show a => [a] -> String
showList [] = "-1"
showList a = unwords [show i | i <- a]

Context

StackExchange Code Review Q#150533, answer score: 7

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