On Fri, Oct 18, 2019 at 06:59:05PM +0200, Ut Primum wrote:
Hi, this is my solution:
productLastPart :: Int -> [Int] -> Int productLastPart n xs = product (take n (reverse xs))
It only uses functions product, take and reverse Regards, Ut
That should work. I’d probably use point-free style, though, and avoid all of those brackets. productLastPart n = product . take n . reverse The version you’ve written runs in linear time, as would a version that worked like this: productLastPart n xs = product (drop (length xs - n) xs) because reverse, length and product (actually a partial application of foldr) are all linear in the length of the list—they access each element once. You can see the GHC implementation of reverse at https://hackage.haskell.org/package/base-4.12.0.0/docs/src/GHC.List.html#rev.... This turns out to be the best we can do with a regular Haskell list, because accessing the last element of a list is a linear-time operation: last :: [a] -> a last [] = error "Oh no!" last [x] = x last (x : xs) = last xs There do, however, exist list types like Sequence and Vector that allow constant-time access to both ends of the list. Hope this is helpful. Seph -- Seph Shewell Brockway, BSc MSc (Glas.) Pronouns: she/her