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copied!<p>With nothing more than a really mindless wrapping of your lists into Vectors I get 2.5 seconds when the input list is <code>[1..10000]</code>.</p> <pre><code>import qualified Data.Vector as V import Data.Vector (Vector, (!)) main = print $ liss [0..10000] liss :: [Int] -> Int liss seqL = V.maximum dp where dp = V.fromList $ 1 : [val n | n <- [1..length seqL - 1]] seq = V.fromList seqL val n = (1 +) . filter'and'get'max ((<= top) . (seq!)) $ [0..pred n] where top = seq!n ----- filter'and'get'max :: (Int -> Bool) -> [Int] -> Int filter'and'get'max f [] = 0 filter'and'get'max f [x] = if f x then dp!x else 0 filter'and'get'max f (x:xs) = if f x then ( if vx > vxs then vx else vxs ) else vxs where vx = dp!x vxs = filter'and'get'max f xs </code></pre> <p>The compilation and execution:</p> <pre><code>tommd@Mavlo:Test$ ghc --version The Glorious Glasgow Haskell Compilation System, version 7.0.3 tommd@Mavlo:Test$ ghc -O2 so.hs [1 of 1] Compiling Main ( so.hs, so.o ) Linking so ... tommd@Mavlo:Test$ time ./so 10001 real 0m2.536s user 0m2.528s </code></pre> <p>A worker-wrapper transformation on <code>filter'and'get'max</code> seems to shave off another second. </p> <p>Also, I don't understand why you need that middle case (<code>filter'and'get'max f [x]</code>), shouldn't it work fine without that? I guess this changes the result if <code>dp!x < 0</code>. Note eliminating that saves 0.3 seconds right there.</p> <p>And the python code you provided takes ~ 10.7 seconds (added a call of <code>longest(range(1,10000));</code>).</p> <pre><code>tommd@Mavlo:Test$ time python so.py real 0m10.745s user 0m10.729s </code></pre>

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