The evident solution is :
unique xs = [x | x <- xs, isIn x xs 2]
isIn _ _ 0 = True isIn _ [] _ = False isIn y (x:xs) n | y == x = isIn y xs (n-1) | otherwise = isIn y xs n
which is quite obviously O(n^2)... (same idea but just a bit faster than Lorenzo one-liner) The solution proposed here is slightly better : for each element, either he is unique and then it is useless to compare the following elements to it, either he isn't and so none of his subsequent occurrences may be unique, we may proceed with the tail of the list without these occurrences. Note that the second solution of franco00 is wrong because it doesn't remove the subsequent occurrences. Still it is O(n^2) as a sum(k from k=n to k=1). Still, we can get better : O(n log n) with a solution based on a sort or Data.Map :
unique xs = nub (sort xs)
-- Jedaï