-- Not beautifully idiomatic, but not too bad, and O(n):
data SolutionState = SSInitial | SS Int Int Int
solve :: [Int] -> SolutionState
solve = foldr go SSInitial where
go x (SS dense best sparse) =
let dense' = max x (dense + x)
best' = max best dense'
sparse' = max (sparse + x) (max sparse x)
in SS dense' best' sparse'
go x SSInitial = SS x x x
On Sat, Jul 16, 2016 at 3:40 PM, Dominik Bollmann wrote: 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
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