Hi all, I've recently been trying to implement the "maximum subarray problem" from [1] in Haskell. My first, naive solution looked like this: maxSubArray :: [Int] -> [Int] maxSubArray [] = [] maxSubArray [x] = [x] maxSubArray xs@(_:_:_) = maxArr (maxArr maxHd maxTl) (maxCrossingArray hd tl) where (hd,tl) = splitAt (length xs `div` 2) xs maxHd = maxSubArray hd maxTl = maxSubArray tl maxCrossingArray :: [Int] -> [Int] -> [Int] maxCrossingArray hd tl | null hd || null tl = error "maxArrayBetween: hd/tl empty!" maxCrossingArray hd tl = maxHd ++ maxTl where maxHd = reverse . foldr1 maxArr . tail $ inits (reverse hd) -- we need to go from the center leftwards, which is why we -- reverse the list `hd'. maxTl = foldr1 maxArr . tail $ inits tl maxArr :: [Int] -> [Int] -> [Int] maxArr xs ys | sum xs > sum ys = xs | otherwise = ys While I originally thought that this should run in O(n*log n), a closer examination revealed that the (++) as well as maxHd and maxTl computations inside function `maxCrossingArray` are O(n^2), which makes solving one of the provided test cases in [1] infeasible. Hence, I rewrote the above code using Data.Array into the following: data ArraySum = ArraySum { from :: Int , to :: Int , value :: Int } deriving (Eq, Show) instance Ord ArraySum where ArraySum _ _ v1 <= ArraySum _ _ v2 = v1 <= v2 maxSubList :: [Int] -> [Int] maxSubList xs = take (to-from+1) . drop (from-1) $ xs where arr = array (1, length xs) [(i,v) | (i,v) <- zip [1..] xs] ArraySum from to val = findMaxArr (1, length xs) arr findMaxArr :: (Int, Int) -> Array Int Int -> ArraySum findMaxArr (start, end) arr | start > end = error "findMaxArr: start > end" | start == end = ArraySum start end (arr ! start) | otherwise = max (max hd tl) (ArraySum leftIdx rightIdx (leftVal+rightVal)) where mid = (start + end) `div` 2 hd = findMaxArr (start, mid) arr tl = findMaxArr (mid+1, end) arr (leftIdx, leftVal) = snd $ findMax mid [mid-1,mid-2..start] (rightIdx, rightVal) = snd $ findMax (mid+1) [mid+2,mid+3..end] findMax pos = foldl' go ((pos, arr ! pos), (pos, arr ! pos)) go ((currIdx, currSum), (maxIdx, maxSum)) idx | newSum >= maxSum = ((idx, newSum), (idx, newSum)) | otherwise = ((idx, newSum), (maxIdx, maxSum)) where newSum = currSum + (arr ! idx) I believe this runs in O(n*log n) now and is fast enough for the purpose of solving the Hackerrank challenge [1]. However, I feel this second solution is not very idiomatic Haskell code and I would prefer the clarity of the first solution over the second, if somehow I could make it more efficient. Therefore my question: What would be an efficient, yet idiomatic solution to solving the "maximum subarray problem" in Haskell? (Note: I'm aware that this problem can be solved in O(n), but I'm also happy with idiomatic Haskell solutions running in O(n*log n)) Thanks, Dominik. [1] https://www.hackerrank.com/challenges/maxsubarray