- Updated solution by Ulrich Rieke.

1 files changed, 26 insertions, 19 deletions

diff --git a/challenge-280/ulrich-rieke/haskell/ch-1.hs b/challenge-280/ulrich-rieke/haskell/ch-1.hs
index 6c48b547ee..346e604a88 100755
--- a/challenge-280/ulrich-rieke/haskell/ch-1.hs
+++ b/challenge-280/ulrich-rieke/haskell/ch-1.hs
@@ -1,19 +1,26 @@
-module Challenge280

-   where

-import qualified Data.Set as S

-import Data.List ( findIndices , sortOn )

-

-solution :: String -> Char

-solution s = if any ( (>= 2) . length . snd ) indices then fst $ head $ 

- sortOn ( last . snd ) $ filter ( (> 1 ) . length . snd ) indices else ' '

- where

-  indices = map (\c -> (c , findIndices (== c ) s ) ) $ S.toList $

-   S.fromList s

-

-main :: IO ( )

-main = do

-   putStrLn "Enter a word of lowercase letters only!"

-   line <- getLine

-   print $ solution line

-

-

+module Challenge280
+   where
+import qualified Data.Set as S
+import Data.List ( findIndices , sortOn , (!!) )
+
+--the solution is as follows : pair up every character with the list
+--of indices where it can be found in the string. We use unique 
+--characters for that. Then , we look for the character that occurs 
+--more than once in the string by filtering all pairs that have more
+--than one element in the associated list. The character we look for 
+--is the one with the smallest index number at position 2 corresponding
+--to index 1 , so we sort on that index
+
+solution :: String -> Char
+solution s = if any ( (>= 2) . length . snd ) indices then fst $ head $ 
+ sortOn (\p -> (snd p) !! 1 ) $ filter ( (> 1 ) . length . snd ) indices
+  else ' '
+  where
+   indices = map (\c -> (c , findIndices (== c ) s ) ) $ S.toList $
+    S.fromList s
+
+main :: IO ( )
+main = do
+   putStrLn "Enter a word of lowercase letters only!"
+   line <- getLine
+   print $ solution line