Flatten a Multilevel Doubly Linked List

Posted on July 15, 2019 by elfsong

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.

Example:

Input:  1---2---3---4---5---6--NULL          
                |          
                7---8---9---10--NULL              
                    |              
                    11--12--NULL 
Output: 1-2-3-7-8-11-12-9-10-4-5-6-NULL

"""
# Definition for a Node.
class Node:
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child
"""

class Solution:
    def flatten(self, head: 'Node') -> 'Node':
        if not head:
            return None
        
        stack = list()
        cur = head
        
        while cur:
            if cur.child:
                if cur.next:
                    stack += [cur.next]
                cur.next = cur.child
                cur.next.prev = cur
                cur.child = None
           
            if not cur.next and stack:
                temp = stack.pop()
                cur.next = temp
                temp.prev = cur
                
            cur = cur.next
        
        return head

Flatten Binary Tree to Linked List

Posted on July 15, 2019 by elfsong

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

The flattened tree should look like:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def flatten(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        node, stack = root, []
        
        while node:
            if node.right:
                stack.append(node.right)
            node.right = node.left
            node.left = None
            
            if not node.right and stack:
                temp = stack.pop()
                node.right = temp
            
            node = node.right

Daily Temperatures

Posted on July 1, 2019 by elfsong

Given a list of daily temperatures T, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.

For example, given the list of temperatures T = [73, 74, 75, 71, 69, 72, 76, 73], your output should be [1, 1, 4, 2, 1, 1, 0, 0].

Note: The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].

Solution:
This quiz is a classical application of the stack data structure. Here we can store indexes of each element which is waiting a bigger element in a stack, and calculate the difference between the current index and stored indexes when the current element is bigger than previous elements. The while loop part is really elegant and tricky, in which judging whether the stack is empty and comparing elements.

class Solution:
    def dailyTemperatures(self, T: List[int]) -> List[int]:
        ans = [0] * len(T)

        stk = list()

        for c_index, value in enumerate(T):
            while stk and T[stk[-1]] < value:
                p_index = stk.pop()
                ans[p_index] = c_index - p_index
            stk.append(c_index)

        return ans

Curiosity

Posted on June 11, 2019 by elfsong

In the world of locked rooms, the man with the key is king. And honey, you should see me in a crown.

Piles of Boxes

Posted on April 11, 2019 by elfsong

Several days ago, I’ve joined a Hackathon that sponsored by Facebook.

As qualifying people to this event, they hand out an easy algorithm question. Shortly, you need to calculate how less time to move a bunch of boxes in same heigh.

Here is a solution for reference:

import collections

def box(boxesInPiles):
    sortBoxes = collections.OrderedDict(sorted(collections.Counter(boxesInPiles).items()))
    count = 0
    for index, key in enumerate(sortBoxes):
        count += index * sortBoxes[key]
    return count

print(box([4, 5, 5, 2, 4]))

Musician

Posted on April 2, 2019 by elfsong

Let’s say there are two persons, Composer and Performer.

The Composer randomly selects three different Pitch which constructed by two part, Note and Octave. Note in the range of “A” to “G”, Octave in the range of “1” to “3”.

For example, here is a typical Pitch combined by ‘A’ and ‘1’, in which ‘A’ is the Note, and ‘1’ is the Octave.

Once Composer selected a Combination, Performer needs to guess it as quick as possible. After each guess, Performer get a feedback which indicates that:

how many pitches in the guess are included in the target (correct pitches)
how many pitches have the right note but the wrong octave (correct notes)
how many pitches have the right octave but the wrong note (correct octaves)

Now, the question is how to get the corrected answer as less times as possible.

The core idea is that when we get a feedback from the previous guess, we’d check out combinations that have identical feedback in the remind possible set, in which the problem scale can be reduced effectively.

Now, the question is how to get the corrected answer as less times as possible.

The core idea is that when we get a feedback from the previous guess, we are able to check out combinations that have identical feedback in the remaining possible set, by which the problem scale can be pruned effectively.

possible set, in which the problem scale can be reduced effectively.

--  Subject  : UniMelb 2019 SM1 COMP90048 Declarative Programming
--  File     : Proj1.hs
--  Author   : Mingzhe Du
--  Origin   : Mon Apr 8 2019 
--  Purpose  : This program for guessing a target chord. In each round of the game,
--             the program will generate a chord from a possible set, and then a feedback 
--             against the guess will be given. Depanding on these feedbacks, the aim of this
--             program is get the correct chord with as less as possible guess times.


module Proj1 (Pitch, GameState, toPitch, feedback, initialGuess, nextGuess) where

-- Pitch structure
data Pitch = Pitch { note :: Char, 
                     octave :: Char   
                   } deriving (Eq)
instance Show Pitch where
    show (Pitch note octave) = [note, octave]

-- Game State
data GameState = GameState { times :: Int,                  -- Guess times
                             cCombinations :: [[[Char]]]    -- Possible set
                           } deriving (Eq, Show)

-- Converting String to Pitch                          
toPitch :: String -> Maybe Pitch
toPitch (a:b:t)
    | not (null t) = Nothing
    | (elem note' ['A'..'G']) && (elem octave' ['1'..'3']) = Just Pitch {note = note', octave = octave'}
    | otherwise = Nothing
    where note' = a
          octave' = b

-- Comparing target chord and guessed chord
feedback :: [Pitch] -> [Pitch] -> (Int,Int,Int)
feedback pitch_a pitch_b
    | (length pitch_a == 3) && (length pitch_b == 3) = (c_p, c_n - c_p, c_o - c_p)
    | otherwise = (0,0,0)
    where get_key key = foldr (\x acc -> key x:acc) []
          c_p = getCounter pitch_a pitch_b 0                              -- Correct pitches
          c_n = getCounter (map note pitch_a) (map note pitch_b) 0        -- Correct notes
          c_o = getCounter (map octave pitch_a) (map octave pitch_b) 0    -- Correct Octaves

-- Initial guess
initialGuess :: ([Pitch], GameState)
initialGuess = (currentGuess, gameState)
    where currentGuess = combinationToPitch  (cGuess)
          gameState = GameState 0 all_combinations 
          all_items = getCombination "ABCDEFG" "123"
          cGuess:all_combinations = subsequencesOfSize 3 all_items        -- New guess and new possible set
          getCombination p_note p_octave = foldr (\x acc -> (map (\y -> y:[x]) p_note) ++ acc) [] p_octave
          combinationToPitch combinations = map (\(Just x) -> x) $ map toPitch combinations     -- Converting String to Pitch

-- Get the next guess
nextGuess :: ([Pitch], GameState) -> (Int,Int,Int) -> ([Pitch],GameState)
nextGuess (pGuess, pGameState) pFeedback = (cGuess, cGameState)
    where cGuess':cCombs = getNewCombination pGuess pCombinations pFeedback
          pCombinations = cCombinations pGameState
          cGuess = toChord cGuess'
          cGameState = GameState ((times pGameState) + 1) cCombs 
          toChord = map (\x -> Pitch (x !! 0) (x !! 1))

-- Help Functions 

-- remove an item from a list
removeItem :: (Eq a) => a -> [a] -> [a]
removeItem _ [] = []
removeItem x (y:ys) 
    | x == y = ys
    | otherwise = y : removeItem x ys 

-- get the number of same elements in two lists
getCounter :: (Eq a) => [a] -> [a] -> Int -> Int
getCounter [] y c = c
getCounter  (x:xs) y c
    | elem x y = getCounter xs (removeItem x y) (c+1)
    | otherwise = getCounter xs y c
 
-- Generate combinations by a specifc size    
subsequencesOfSize :: Int -> [a] -> [[a]]
subsequencesOfSize n xs = let l = length xs in if n>l then [] else subsequencesBySize xs !! (l-n)
    where subsequencesBySize [] = [[[]]]
          subsequencesBySize (x:xs) = let next = subsequencesBySize xs in zipWith (++) ([]:next) (map (map (x:)) next ++ [[]])

-- Converting a string list to pitch list
toChord :: [[Char]] -> [Pitch]
toChord a = map (\x -> Pitch (x !! 0) (x !! 1)) a 

-- retrive a new guess
getNewCombination :: [Pitch] -> [[[Char]]] -> (Int, Int, Int) -> [[[Char]]]
getNewCombination guess allCombinations pFeedback = foldr (\x acc -> if checkFeedback (toChord x) then  x:acc else acc) [] allCombinations
    where checkFeedback nChord = if pFeedback == (feedback guess nChord) then True else False

Haskell moment

Posted on March 28, 2019 by elfsong

Here are some Haskell code chunks, which both are simple recursion algorithms.

Quicksort is a sort of poster child for Haskell because everyone does it to showcase how elegant Haskell is.

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []
quicksort (x:xs) =
let smallerSorted = quicksort [a | a <- xs, a <= x]
    biggerSoted = quicksort [a | a <- xs, a > x]
in smallerSorted ++ [x] ++ biggerSoted

maximum' :: (Ord a) => [a] -> a
maximum' [] = error "List is empty!"
maximum' [x] = x
maximum' (x:xs) = (max x (maximum' xs))

replicate' :: Int  -> b -> [b]
replicate' n x
    | n <= 0 = []
    | otherwise = (x:(replicate' (n-1) x))

take' :: Int -> [a] -> [a]
take' n (x:xs)
    | n <= 0 = []
    | otherwise = x:(take' (n-1) xs)

reserve' :: [a] -> [a]
reserve' [] = []
reserve' (x:xs) = (reserve' xs) ++ [x]