Implement solution in Haskell as depth-first-search

src/l33tcode/841.hs

@@ -0,0 +1,52 @@
+{-
+Keys and Rooms
+
+Instructions:
+There are n rooms labelled from 0 to n - 1 and all the rooms are locked except for room 0. Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key.
+
+When you visit a room, you may find a set of distinct keys in it. Each key has a number on it, denoting which room it unlocks, and you can take all of them with you to unlock the other rooms.
+
+Given an array rooms where rooms[i] is the set of keys that you can obtain if you visited room i, return true if you can visit all the rooms, or false otherwise.
+
+Constraints:
+    n == rooms.length
+    2 <= n <= 1000
+    0 <= rooms[i].length <= 1000
+    1 <= sum(rooms[i].length) <= 3000
+    0 <= rooms[i][j] < n
+    All the values of rooms[i] are unique.
+-}
+module Main where
+
+import           Data.List (genericIndex, genericLength)
+
+input1Rooms :: [[Integer]]
+input1Rooms = [[1],[2],[3],[]]
+
+input2Rooms :: [[Integer]]
+input2Rooms = [[1,3],[3,0,1],[2],[0]]
+
+main :: IO ()
+main = do
+  print $ "Example 1 is: " <> show (canVisitAll input1Rooms)
+  print $ "Example 2 is: " <> show (canVisitAll input2Rooms)
+
+canVisitAll :: [[Integer]] -> Bool
+canVisitAll rooms = [0..l] == dfs rooms 0 []
+  where
+    l = l' rooms - 1
+
+dfs :: [[Integer]] -> Integer -> [Integer] -> [Integer]
+dfs graph current visited =
+    foldl (\visited next ->
+              if next `elem` visited
+                then visited
+                else dfs graph next visited)
+           (visited ++ [current])
+           (graph !!! current)
+
+l' :: [a] -> Integer
+l' = genericLength
+
+(!!!) :: [a] -> Integer -> a
+(!!!) = genericIndex