Monthly Archives: September 2019

Longest Consecutive Sequence

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Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

Your algorithm should run in O(n) complexity.

Example:

Input: [100, 4, 200, 1, 3, 2]
Output: 4
Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        nums = set(nums)
        ans = 0
        
        for x in nums:
            # Find the first element in one segment
            if x-1 not in nums:
                y = x + 1
                # reach the last consecutive element in one element  
                while y in nums:
                    y += 1
                ans = max(ans, y - x)
             
        return ans
class Solution:
    def longestConsecutive(self, nums) -> int:
        UF = {}
        nums = set(nums)

        if not nums:
            return 0

        # for num in nums:
        #     UF[num] = num

        def find(x):
            if x != UF[x]:
                UF[x] = find(UF[x])
            return UF[x]

        def union(x, y):
            UF.setdefault(x, x)
            UF.setdefault(y, y)
            UF[find(x)] = find(y)

        for n in nums:
            if (n - 1) in nums:
                union(n - 1, n)
            if (n + 1) in nums:
                union(n, n + 1)

        ans = 1

        for num in nums:
            if num in UF:
                ans = max(ans, find(num) - num + 1)
        return ans

Python Daemon

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# coding=utf8
import os
import sys
import atexit


def daemonize(pid_file=None):
    """
    Create daemon process
    :param pid_file: File that saves process id
    :return:
    """
    # Fork a subprocess from parent process
    pid = os.fork()

    # The id of subprocess equals 0, and of parent process must larger than 0
    if pid:
        # Exit from parent process(sys.exit() executes flush while os._exit() doesn't)
        sys.exit(0)

    # The subprocess defaults the working directory of its parent process. 
    os.chdir('/')

    # The subprocess defaults umask(File permission mask), reset it to 0(Fully control) in order to avoiding R/W operations.
    os.umask(0)

    # Let the subprocess be the leader of the conversation and the process.
    os.setsid()

    # Second Fork
    _pid = os.fork()
    if _pid:
        # Exit subprocess
        sys.exit(0)

    # Grandson process is daemon right now. Redirecting stdin/stdout/stderr descriptions to avoid errors in the printing process.

    # Flush stdout/stderr buffer
    sys.stdout.flush()
    sys.stderr.flush()

    # Atomically close and duclipate file description, and redirect into /dev/null(Drop off all input and output)
    with open('/dev/null') as read_null, open('/dev/null', 'w') as write_null:
        os.dup2(read_null.fileno(), sys.stdin.fileno())
        os.dup2(write_null.fileno(), sys.stdout.fileno())
        os.dup2(write_null.fileno(), sys.stderr.fileno())

    # Write PID
    if pid_file:
        with open(pid_file, 'w+') as f:
            f.write(str(os.getpid()))
        # Register Exit function for abnormal exit
        atexit.register(os.remove, pid_file)
  1. Fork sub-process, and exit parent-process.
  2. Change Working Directory (chdir), File Permission Mask(umask), Process Group and Conversation Group (setsid).
  3. Fork grandson-process, and exit sub-process.
  4. Flush buffer and atomically close and duclipate file description, and redirect into /dev/null(Drop off all input and output)
  5. (Option) Write PID.

Most Stones Removed with Same Row or Column

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On a 2D plane, we place stones at some integer coordinate points.  Each coordinate point may have at most one stone.

Now, a move consists of removing a stone that shares a column or row with another stone on the grid.

What is the largest possible number of moves we can make?

Example 1:

Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5

Example 2:

Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3

Example 3:

Input: stones = [[0,0]]

At very first I read this problem, the MOVE operation is described as confusing as heck. I read some explanations on the discussboard, and then finally got the actual meaning.

We can treat each coordinate as two parts (x and y), and can maintain an Union-Find in order to calculate how many groups in the entire field.

 class Solution:
    def removeStones(self, stones: List[List[int]]) -> int:
        UF = {}
        
        def find(x):
            if x != UF[x]:
                UF[x] = find(UF[x])
            return UF[x]
        
        def union(x, y):
            UF.setdefault(x, x)
            UF.setdefault(y, y)
            UF[find(x)] = find(y)


        for i, j in stones:
            print(i, j, ~j)
            union(i, ~j)
            
        return len(stones) - len({find(x) for x in UF})

Brace Expansion

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A string S represents a list of words.

Each letter in the word has 1 or more options.  If there is one option, the letter is represented as is.  If there is more than one option, then curly braces delimit the options.  For example, "{a,b,c}" represents options ["a", "b", "c"].

For example, "{a,b,c}d{e,f}" represents the list ["ade", "adf", "bde", "bdf", "cde", "cdf"].

Return all words that can be formed in this manner, in lexicographical order.

Example 1:

Input: "{a,b}c{d,e}f"
Output: ["acdf","acef","bcdf","bcef"]

Example 2:

Input: "abcd"
Output: ["abcd"]

Note:

  1. 1 <= S.length <= 50
  2. There are no nested curly brackets.
  3. All characters inside a pair of consecutive opening and ending curly brackets are different.
class Solution:
    def expand(self, S):

        def func(remain, result, results):
            if not remain:
                results.append(result)

            else:
                # if a single character appear at the first of the remain string
                if remain[0] != "{":
                    func(remain[1:], result + remain[0], results)
                    return
                else:
                    temp = list()
                    i = 0

                    # find elements in the first brace of the remain string
                    while remain[i] != "}":
                        if remain[i].isalpha():
                            temp.append(remain[i])
                        i += 1

                    for t in temp:
                        func(remain[i + 1:], result + t, results)

        results = []

        func(S, "", results)

        return results

In me the tiger sniffs the rose

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I wanted you to see what real courage is, instead of getting the idea that courage is a man with a gun in his hand. It’s when you know you’re licked before you begin anyway and you see it through no matter what. You rarely win, but sometimes you do.

我想让你见识一下什么事真正的勇气。勇气并不是一个人手里拿着枪，而是当你还未开始就已知道自己会输，可你依然要去做，而且无论如何都要把它坚持到底。你很少能赢，但有时也会。

Delete Nodes And Return Forest

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Given the root of a binary tree, each node in the tree has a distinct value.

After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees).

Return the roots of the trees in the remaining forest.  You may return the result in any order.

Example 1:

Input: root = [1,2,3,4,5,6,7], to_delete = [3,5]
Output: [[1,2,null,4],[6],[7]]

Constraints:

  • The number of nodes in the given tree is at most 1000.
  • Each node has a distinct value between 1 and 1000.
  • to_delete.length <= 1000
  • to_delete contains distinct values between 1 and 1000.
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def delNodes(self, root: TreeNode, to_delete: List[int]) -> List[TreeNode]:
        forest = []
        to_delete = set(to_delete)
        
        def buttom_up(root):
            if root:
                left, right = dfs(root.left), dfs(root.right)
                
                root.left, root.right = left, right
                
                if root.val in to_delete:
                    if left: forest.append(left)
                    if right: forest.append(right)
                else:
                    return root
        
        dfs(root)
        
        return ([] if root.val in to_delete else [root]) + forest
        
        #########################################################
        
        to_delete_set = set(to_delete)
        res = []

        def top_down(root, is_root):
            if not root: 
                return None
            
            root_deleted = root.val in to_delete_set
            
            if is_root and not root_deleted:
                res.append(root)
                
            root.left = helper(root.left, root_deleted)
            root.right = helper(root.right, root_deleted)
            
            return None if root_deleted else root
        
        top_down(root, True)
        
        return res

Decode String

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Given an encoded string, return its decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won’t be input like 3a or 2[4].

Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
class Solution:
    def decodeString(self, s: str) -> str:
        stack = list()
        times = list()

        for i, v in enumerate(s):
            if v.isdigit():
                # Single Digit, like 1,2,3...
                if not s[i-1].isdigit():
                    times.append(int(v))
                # Multiple digits, like 100, 200...
                else:
                    times[-1] = times[-1] * 10 + int(v)
            
            # push
            elif v != "]":
                stack.append(v)
            
            # reach "]"
            else:
                # Retrive string in cloest level square brackets
                b = ""
                t = stack.pop()
                while t != "[":
                    b = t + b
                    t = stack.pop()

                stack.append(b * times.pop())

        return "".join(stack)

Sentence Similarity II

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Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar.

For example, words1 = ["great", "acting", "skills"] and words2 = ["fine", "drama", "talent"] are similar, if the similar word pairs are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]].

Note that the similarity relation is transitive. For example, if “great” and “good” are similar, and “fine” and “good” are similar, then “great” and “fine” are similar.

Similarity is also symmetric. For example, “great” and “fine” being similar is the same as “fine” and “great” being similar.

Also, a word is always similar with itself. For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though there are no specified similar word pairs.

Finally, sentences can only be similar if they have the same number of words. So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].

Note:

  • The length of words1 and words2 will not exceed 1000.
  • The length of pairs will not exceed 2000.
  • The length of each pairs[i] will be 2.
  • The length of each words[i] and pairs[i][j] will be in the range [1, 20].
class Solution:
    def areSentencesSimilarTwo(self, words1: List[str], words2: List[str], pairs: List[List[str]]) -> bool:
        parent = {}
        
        if len(words1) != len(words2):
            return False
        
        def find(w):
            if w not in parent:
                return w
            if parent[w] != w:
                w = find(parent[w])
            return w
        
        
        def union(w1, w2):
            parent[w1] = w2
        
        
        for pair in pairs:
            union(find(pair[0]), find(pair[1]))
        
        
        for word_pair in zip(words1, words2):
            if word_pair[0] == word_pair[1]:
                continue
            else:
                leader1 = find(word_pair[0])
                leader2 = find(word_pair[1])
                if leader1 == leader2:
                    continue
                else:
                    return False
        return True

Campus Bikes II

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On a campus represented as a 2D grid, there are N workers and M bikes, with N <= M. Each worker and bike is a 2D coordinate on this grid.

We assign one unique bike to each worker so that the sum of the Manhattan distances between each worker and their assigned bike is minimized.

The Manhattan distance between two points p1 and p2 is Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|.

Return the minimum possible sum of Manhattan distances between each worker and their assigned bike.

Example 1:

Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]]
Output: 6
Explanation: 
We assign bike 0 to worker 0, bike 1 to worker 1. The Manhattan distance of both assignments is 3, so the output is 6.

Example 2:

Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]]
Output: 4
Explanation: 
We first assign bike 0 to worker 0, then assign bike 1 to worker 1 or worker 2, bike 2 to worker 2 or worker 1. Both assignments lead to sum of the Manhattan distances as 4.

Note:

  1. 0 <= workers[i][0], workers[i][1], bikes[i][0], bikes[i][1] < 1000
  2. All worker and bike locations are distinct.
  3. 1 <= workers.length <= bikes.length <= 10
from functools import lru_cache

class Solution:
    def dis(self, a, b):
        return abs(a[0] - b[0]) + abs(a[1] - b[1])
        
    def assignBikes(self, workers: List[List[int]], bikes: List[List[int]]) -> int:
        @lru_cache(None)
        def dfs(p, arr):
            return 0 if p == self.W else min([self.dis(bikes[i], workers[p]) + dfs(p + 1, arr + 2 ** i) for i in range(self.B) if not (arr>>i) & 1])

        self.W = len(workers)
        self.B = len(bikes)
        
        return dfs(0, 0)