CODEWITHSUNDEEP
sortinghaskell
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Reputation: 859

How to find a second largest number in a list in Haskell?

The question is like this:

Write a function that takes an integer list and returns its length and the second largest integer in the list.

I can solve this with two functions but is there any solution that use only one function to do it?

Any help is appreciated! Thanks!

Upvotes: 2

Views: 3945

Answers (4)

klapaucius
klapaucius

Reputation: 1102

head . (!!1) . group . sortBy (flip compare) $ [1,1,2,3,4,4,4,5,5,6,6,6,6]
5

Upvotes: 1

Daniel Wagner
Daniel Wagner

Reputation: 152707

Don't make it complicated.

f xs = (sort xs !! 1, length xs)

If you choose to make it complicated, make it complicated in the right way.

Upvotes: 4

bheklilr
bheklilr

Reputation: 54058

Edited to use @ThomasM.DuBuisson's suggestion

You can solve this the same way that you could finding the max: using a fold. Max can be pretty trivially implemented with

mymaximum :: Ord a => [a] -> a
mymaximum xs = foldl searcher (head xs) xs
    where
        searcher :: Ord a => a -> a -> a
        searcher a b
            | a > b     = a
            | otherwise = b

So we can implement it similarly by just keeping up with the two largest values (notice that we have to "seed" the fold with a tuple as well):

nextmaximum :: Ord a => [a] -> a
nextmaximum xs = fst $ foldl searcher (h, h) xs
    where
        h = head xs
        searcher :: (Ord a) => (a, a) -> a -> (a, a)
        searcher (s, f) x = (min f (max s x), max f x)

Upvotes: 3

jtobin
jtobin

Reputation: 3273

You can just compose individual functions together. This is neither efficient nor robust, but it's sure easy to write:

f xs = maximum . filter (< maximum xs) $ xs

Upvotes: 2

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