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question:# Advanced Problem: Lowest Common Ancestor in a Binary Tree Scenario As a software engineer involved in the development of a large-scale application, you need to implement a feature that determines the lowest common ancestor (LCA) of two nodes in a binary tree. This is a fundamental task in various applications, such as network routing algorithms and genealogy software. Requirements * **Function Signature**: Implement the function `find_lca(root: TreeNode, node1: TreeNode, node2: TreeNode) -> TreeNode`. * **Input**: - `root` (TreeNode): The root of the binary tree. - `node1` (TreeNode): The first node for which the LCA is being calculated. - `node2` (TreeNode): The second node for which the LCA is being calculated. * **Output**: - Return the TreeNode that represents the lowest common ancestor of `node1` and `node2`. Constraints 1. Each node in the binary tree contains a unique integer value. 2. Both `node1` and `node2` are guaranteed to exist in the binary tree. 3. The binary tree can have up to `10^5` nodes. 4. TreeNode class is defined with: ```python class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None ``` Performance Requirements - The solution must handle large inputs efficiently. - Aim for a time complexity of O(n), where n is the number of nodes in the tree. Example ```python class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def find_lca(root: TreeNode, node1: TreeNode, node2: TreeNode) -> TreeNode: if not root: return None if root == node1 or root == node2: return root left = find_lca(root.left, node1, node2) right = find_lca(root.right, node1, node2) if left and right: return root return left if left else right # Example usage: # Constructing the binary tree: # 3 # / # 5 1 # / / # 6 2 0 8 # / # 7 4 root = TreeNode(3) root.left = TreeNode(5) root.right = TreeNode(1) root.left.left = TreeNode(6) root.left.right = TreeNode(2) root.right.left = TreeNode(0) root.right.right = TreeNode(8) root.left.right.left = TreeNode(7) root.left.right.right = TreeNode(4) node1 = root.left # Node with value 5 node2 = root.left.right.right # Node with value 4 # Expected output: Node with value 5 print(find_lca(root, node1, node2).val) ``` Ensure that your implementation passes various edge cases such as: 1. When the nodes are at different depths. 2. When one node is the ancestor of the other. 3. When the two nodes are in different subtrees.

answer:class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None def find_lca(root: TreeNode, node1: TreeNode, node2: TreeNode) -> TreeNode: if not root: return None if root == node1 or root == node2: return root left_lca = find_lca(root.left, node1, node2) right_lca = find_lca(root.right, node1, node2) if left_lca and right_lca: return root return left_lca if left_lca else right_lca

question:# Context You are tasked with implementing a simplified version of the `set` data structure in Python that supports basic operations such as insertion, deletion, membership checking, and union/intersection of two sets. Your implementation should mimic the behavior of Python's built-in `set`. # Problem Statement Implement a `SimpleSet` class that supports the following methods: - `add(self, element)`: Adds the element to the set if it is not already present. - `remove(self, element)`: Removes the element from the set. Raises a `KeyError` if the element is not found. - `contains(self, element)`: Returns `True` if the element is in the set, `False` otherwise. - `__len__(self)`: Returns the number of elements in the set. - `union(self, other_set)`: Returns a new `SimpleSet` that is the union of the current set and `other_set`. - `intersection(self, other_set)`: Returns a new `SimpleSet` that is the intersection of the current set and `other_set`. # Requirements 1. Your `SimpleSet` must use a hash table to store elements to ensure efficient operations. 2. Handle hash collisions using chaining (linked lists). 3. The initial capacity of the hash table should be 8. 4. Dynamically resize the table when the load factor exceeds 0.75. # Constraints - Your implementation should aim for an average time complexity of O(1) for add, remove, and contains operations. - Use Python's built-in `hash` function for hashing elements. - Elements will not be None and will be hashable. # Input/Output - There is no direct input/output. Implement the class and methods as specified below: ```python class SimpleSet: def __init__(self, initial_capacity=8): # Initialize your data structure pass def add(self, element): # Add element to the set pass def remove(self, element): # Remove element from the set pass def contains(self, element): # Check if the element exists in the set pass def __len__(self): # Return the number of elements in the set pass def union(self, other_set): # Return a new SimpleSet containing elements from both sets pass def intersection(self, other_set): # Return a new SimpleSet containing elements common to both sets pass ``` # Example Usage ```python set1 = SimpleSet() set1.add("a") set1.add("b") set1.add("c") set2 = SimpleSet() set2.add("b") set2.add("c") set2.add("d") print(set1.contains("a")) # Output: True print(set1.contains("d")) # Output: False set1.remove("a") print(len(set1)) # Output: 2 union_set = set1.union(set2) print(len(union_set)) # Output: 4 intersection_set = set1.intersection(set2) print(len(intersection_set)) # Output: 2 print(intersection_set.contains("b")) # Output: True ``` This question is designed to be consistent with the style, length, difficulty level, and topic alignment of the sample question provided. It focuses on implementing a basic data structure with essential methods and requiring consideration of efficiency and proper handling of hash collisions.

answer:class SimpleSet: def __init__(self, initial_capacity=8): self.capacity = initial_capacity self.size = 0 self.buckets = [[] for _ in range(self.capacity)] def _hash(self, element): return hash(element) % self.capacity def _resize(self): new_capacity = self.capacity * 2 new_buckets = [[] for _ in range(new_capacity)] for bucket in self.buckets: for element in bucket: index = hash(element) % new_capacity new_buckets[index].append(element) self.capacity = new_capacity self.buckets = new_buckets def add(self, element): if not self.contains(element): if self.size / self.capacity > 0.75: self._resize() index = self._hash(element) self.buckets[index].append(element) self.size += 1 def remove(self, element): index = self._hash(element) bucket = self.buckets[index] if element in bucket: bucket.remove(element) self.size -= 1 else: raise KeyError(f"{element} not found in the set") def contains(self, element): index = self._hash(element) bucket = self.buckets[index] return element in bucket def __len__(self): return self.size def union(self, other_set): result_set = SimpleSet() for bucket in self.buckets: for element in bucket: result_set.add(element) for bucket in other_set.buckets: for element in bucket: result_set.add(element) return result_set def intersection(self, other_set): result_set = SimpleSet() for bucket in self.buckets: for element in bucket: if other_set.contains(element): result_set.add(element) return result_set

question:# Question Given a non-negative integer `n`, implement a function that returns the factorial of `n`. The factorial of a non-negative integer `n` is the product of all positive integers less than or equal to `n` and is denoted by `n!`. Your task is to write a function `factorial` that takes an integer as input and returns its factorial. Function Signature ```python def factorial(n: int) -> int: pass ``` Input - An integer `n` (0 <= n <= 20). Output - An integer representing the factorial of `n`. Constraints - The input integer `n` will be within the range of 0 to 20, ensuring the result can be held in a standard integer type without overflow. Example ```python # Example 1 n = 5 print(factorial(n)) # Output: 120 # Example 2 n = 0 print(factorial(n)) # Output: 1 # Example 3 n = 3 print(factorial(n)) # Output: 6 ``` Additional Notes - Consider edge cases such as the smallest input (n = 0) and the largest input (n = 20). - Ensure that your implementation is efficient and correctly handles the input constraints. - Write clean and readable code with comments to explain the recursive or iterative approach used.

answer:def factorial(n: int) -> int: Returns the factorial of a non-negative integer n. Args: n (int): A non-negative integer whose factorial is to be computed. Returns: int: The factorial of the input integer n. if n == 0: return 1 # By definition, 0! is 1 result = 1 for i in range(1, n + 1): result *= i return result

question:# Movie Ranking System You have been asked to create a function that generates rankings for a list of movies based on their ratings. Each movie has a unique identifier and a rating score. The function should return a list of tuples, where each tuple contains the movie identifier and its relative rank. Function Signature ```python def rank_movies(movies: List[Tuple[str, float]]) -> List[Tuple[str, int]]: Ranks the given list of movies based on their ratings. Parameters: - movies (List[Tuple[str, float]]): A list of tuples where each tuple contains: - a string representing the movie identifier (unique name or ID), - a float representing the movie's rating (a value between 0.0 and 10.0). Returns: - List[Tuple[str, int]]: A list of tuples where each tuple contains: - the movie identifier, - the rank of the movie (integer) where 1 is the highest rank. Movies with the same rating should receive the same rank. The next rank after a tie will skip the number of tied movies to maintain sequential ranking. ``` Additional Constraints and Information: 1. **Input Validation**: - Each movie identifier must be a non-empty string. - Each rating must be a float value in the range 0.0 to 10.0. - The input list must contain at least one movie. 2. **Edge Cases**: - Multiple movies with the same rating (ties). - Invalid movie identifier (empty string). - Invalid rating (out of range). 3. **Ranking Format**: - The highest rating should be ranked as 1. - Movies with the same rating should share the same rank, and there should be a gap before the next rank. Examples: ```python >>> rank_movies([("Movie A", 8.5), ("Movie B", 9.2), ("Movie C", 7.8), ("Movie D", 9.2)]) [('Movie B', 1), ('Movie D', 1), ('Movie A', 3), ('Movie C', 4)] >>> rank_movies([("Movie X", 5.5), ("Movie Y", 5.5), ("Movie Z", 9.0)]) [('Movie Z', 1), ('Movie X', 2), ('Movie Y', 2)] >>> rank_movies([("Alpha", 10.0)]) [('Alpha', 1)] >>> rank_movies([]) Traceback (most recent call last): ... ValueError: The input list must contain at least one movie. >>> rank_movies([("Invalid", 11.0)]) Traceback (most recent call last): ... ValueError: Rating must be a float between 0.0 and 10.0. >>> rank_movies([("", 5.0)]) Traceback (most recent call last): ... ValueError: Movie identifier must be a non-empty string. ``` Write the `rank_movies` function in Python as per the described requirements.

answer:from typing import List, Tuple def rank_movies(movies: List[Tuple[str, float]]) -> List[Tuple[str, int]]: Ranks the given list of movies based on their ratings. Parameters: - movies (List[Tuple[str, float]]): A list of tuples where each tuple contains: - a string representing the movie identifier (unique name or ID), - a float representing the movie's rating (a value between 0.0 and 10.0). Returns: - List[Tuple[str, int]]: A list of tuples where each tuple contains: - the movie identifier, - the rank of the movie (integer) where 1 is the highest rank. if not movies: raise ValueError("The input list must contain at least one movie.") for movie_id, rating in movies: if not isinstance(movie_id, str) or not movie_id: raise ValueError("Movie identifier must be a non-empty string.") if not (isinstance(rating, float) and 0.0 <= rating <= 10.0): raise ValueError("Rating must be a float between 0.0 and 10.0.") # Sort movies by rating in descending order, the highest rating first movies_sorted = sorted(movies, key=lambda x: (-x[1], x[0])) result = [] current_rank = 1 for i, (movie_id, rating) in enumerate(movies_sorted): if i > 0 and movies_sorted[i-1][1] != rating: current_rank = i + 1 result.append((movie_id, current_rank)) return result