Category Archives: Leetcode

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

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)

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, to_delete):
        """
        :type root: TreeNode
        :type to_delete: List[int]
        :rtype: List[TreeNode]
        """
        self.forests = []
        self.delete = to_delete
        
        node = self.divideAndConquer(root)
        
        if node:
            self.forests.append(node)
        
        return self.forests
    
    def divideAndConquer(self, node):
        
        if not node:
            return None
        
        left = self.divideAndConquer(node.left)
        right = self.divideAndConquer(node.right)
        
        if node.val not in self.delete:
            node.left = left
            node.right = right
            return node
        else:
            if left:
                self.forests.append(left)
            if right:
                self.forests.append(right)
                
            return None

Minimum Knight Moves

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In an infinite chess board with coordinates from -infinity to +infinity, you have a knight at square [0, 0].

A knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.

Return the minimum number of steps needed to move the knight to the square [x, y].  It is guaranteed the answer exists.

Example 1:

Input: x = 2, y = 1
Output: 1
Explanation: [0, 0] → [2, 1]

Example 2:

Input: x = 5, y = 5
Output: 4
Explanation: [0, 0] → [2, 1] → [4, 2] → [3, 4] → [5, 5]
class Solution:
    def minKnightMoves(self, x: int, y: int) -> int:
        x, y = abs(x), abs(y)
        
        q = collections.deque([(0,0,0)])
        seen = set([(0,0)])

        while q:
            s_x, s_y, s = q.popleft()
            if (s_x, s_y) == (x,y):
                return s

            d = []
            for dx, dy in [(-2,-1),(-2,1),(-1,2),(1,2),(2,1),(2,-1),(1,-2),(-1,-2)]:
                n_x = s_x + dx
                n_y = s_y + dy
                if -2 <= n_x <= x and -2 <= n_y <= y:
                    d.append((n_x,n_y))

            d.sort(key=lambda e: abs(x-e[0]) + abs(y-e[1]))

			# only enqueue towards the 2 directions closest to the x,y
            for n_x,n_y in d[:2]:
                if (n_x, n_y) not in seen:
                    q.append((n_x,n_y,s+1))
                    seen.add((n_x,n_y))

Sum Root to Leaf Numbers

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Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sumNumbers(self, root: TreeNode) -> int:
        def getNumbers(root, previous, result):
            if not root:
                return
            
            current = previous*10 + root.val
            
            if not root.left and not root.right:
                result[0] += current
            
            getNumbers(root.left, current, result)
            getNumbers(root.right, current, result)
            
        ans = [0]
        getNumbers(root, 0, ans)
        return ans[0]

Decode ways

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A message containing letters from A-Z is being encoded to numbers using the following mapping:

'A' -> 1
'B' -> 2
...
'Z' -> 26

Given a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

Example 2:

Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
class Solution:
    def numDecodings(self, s: str) -> int:        
        if s[0] == '0': return 0
        
#         N = len(s)
#         dp = [1 for i in range(N+1)]
        
        for i in range(1, len(s)):
            if s[i] == '0':
                # dp[i] = 0
                b = 0
            if int(s[i-1:i+1]) <= 26:
                # dp[i+1] = dp[i] + dp[i-1]
                c = b+a
            else:
                # dp[i+1] = dp[i]
                c = b
            a=b
            b=c
            
        return dp[-1]