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question:# Binary Tree to Doubly Linked List Conversion You are given a binary tree with the following node structure: ```python class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right ``` Write a function `convert_bin_tree_to_dll` that converts the given binary tree into a doubly linked list, using an in-order traversal. The left pointer in the tree node should be used as the previous pointer in the doubly linked list, and the right pointer should be used as the next pointer. Your function should return the head of the doubly linked list. # Function Signature ```python def convert_bin_tree_to_dll(root: TreeNode) -> TreeNode: ``` # Input * `root`: The root of a binary tree (`TreeNode`). # Output * The head of the doubly linked list (`TreeNode`). # Constraints * The number of nodes in the tree is in the range `[0, 1000]`. * `-10^5 <= Node.val <= 10^5` # Requirements * The solution should perform the transformation efficiently with respect to time and space. * The resulting linked list should represent an in-order traversal of the tree. # Example Example 1: ``` Input: 4 / 2 5 / 1 3 Output: 1 <-> 2 <-> 3 <-> 4 <-> 5 ``` Example 2: ``` Input: 1 / 2 3 Output: 2 <-> 1 <-> 3 ``` # Edge Cases * Empty tree: `root` is None. * Single node tree. * Left-skewed or right-skewed trees. # Notes * Use recursion or an iterative approach to achieve the in-order traversal and transformation. * Make sure to handle edge cases and null nodes appropriately to avoid any unintended errors.
answer:class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def convert_bin_tree_to_dll(root: TreeNode) -> TreeNode: Function to convert a binary tree to a doubly linked list using in-order traversal. Uses the left pointer as the previous and the right as the next pointer. Returns the head of the doubly linked list. if not root: return None # Helper function to convert the tree def helper(node): nonlocal last, head if node: # Recursively visit left subtree helper(node.left) # Processing the current node if last: # Link the previous node (last) with the current one (node) last.right = node node.left = last else: # If this is the leftmost node, it will be the head of the list head = node last = node # Update last to the current node # Recursively visit right subtree helper(node.right) last = None # This will keep track of the last processed node in the DLL head = None # This will be the head of our DLL helper(root) return head
question:# Context A Segment Tree is a powerful data structure that supports fast range queries and point updates. Given an array, you are to implement a Segment Tree class which uses a commutative function provided during initialization to process range queries and point updates efficiently in logarithmic time. # Task Implement a class `SegmentTree` that supports the following functionalities: 1. **Initialization**: Initialize the Segment Tree with an array and a commutative function. 2. **Point Update**: Update the value at a specific index. 3. **Range Query**: Query the result of the commutative function applied over a range of indices. # Details * **Function Signature**: ```python class SegmentTree: def __init__(self, arr: List[Any], function: Callable[[Any, Any], Any]): # Initialize your segment tree pass def update(self, index: int, value: Any): # Update the element at index with the new value pass def query(self, left: int, right: int) -> Any: # Return the result of the function applied from index left to right pass ``` * **Input/Output**: * `__init__(arr: List[Any], function: Callable[[Any, Any], Any])`: * `arr`: An array of any data type. * `function`: A commutative function taking two inputs of the array's data type. * `update(index: int, value: Any)`: No output. Updates the value at `index` in the array. * `query(left: int, right: int) -> Any`: Returns the result of the function applied over the range `[left,right]` inclusive. * **Constraints**: * 1 ≤ len(arr) ≤ 10⁵ * 0 ≤ index, left, right < len(arr) * The commutative function is guaranteed to return results within a reasonable bound. # Example Usage ```python # Example with maximum function arr = [2, 4, 5, 3, 4] mytree = SegmentTree(arr, max) print(mytree.query(2, 4)) # Output: 5 mytree.update(3, 6) print(mytree.query(0, 3)) # Output: 6 # Example with summation function using lambda arr = [4, 5, 2, 3, 4, 43, 3] mytree = SegmentTree(arr, lambda a, b: a + b) print(mytree.query(0, 6)) # Output: 64 mytree.update(2, -10) print(mytree.query(0, 6)) # Output: 52 ```
answer:from typing import List, Callable, Any class SegmentTree: def __init__(self, arr: List[Any], function: Callable[[Any, Any], Any]): self.n = len(arr) self.function = function self.tree = [None] * (2 * self.n) # Tree array to store segment tree # Build tree by inserting leaf elements in tree array for i in range(self.n): self.tree[self.n + i] = arr[i] # Build the tree by calculating parents for i in range(self.n - 1, 0, -1): self.tree[i] = self.function(self.tree[2 * i], self.tree[2 * i + 1]) def update(self, index: int, value: Any): # Set value at position p pos = index + self.n self.tree[pos] = value # Then move upward and update nodes i = pos while i > 1: i //= 2 self.tree[i] = self.function(self.tree[2 * i], self.tree[2 * i + 1]) def query(self, left: int, right: int) -> Any: # Get the sum of elements between indices l and r inclusive res = None l = left + self.n r = right + self.n + 1 while l < r: if l % 2 == 1: if res is None: res = self.tree[l] else: res = self.function(res, self.tree[l]) l += 1 if r % 2 == 1: r -= 1 if res is None: res = self.tree[r] else: res = self.function(res, self.tree[r]) l //= 2 r //= 2 return res
question:# Context Anagram detection is a common problem in text processing and cryptography. An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once. Given two strings consisting of lowercase English letters, your task is to determine if the two strings are anagrams of each other. # Task Write a function `are_anagrams` that takes two strings as input and returns `True` if they are anagrams of each other and `False` otherwise. # Input/Output Format Input * Two strings, `s1` and `s2`. (1<= len(s1), len(s2) <= 10^5) Output * Boolean value - `True` if the strings are anagrams, `False` otherwise. # Constraints * The input strings will consist of lowercase English letters only. * You cannot use any sorting mechanisms. # Performance Requirements * The function should operate in O(n) time complexity where n is the length of the strings. * Space complexity should be O(1) since the frequency array is constant and does not depend on the input size. # Implementation ```python def are_anagrams(s1, s2): # Your implementation goes here pass # Example usage: print(are_anagrams("apple", "pleap")) # True print(are_anagrams("apple", "cherry")) # False ``` # Explanation 1. Initialize two arrays of size 26 (to count character frequencies for `s1` and `s2`). 2. Traverse both strings and update the corresponding frequency arrays. 3. Compare the frequency arrays; if they are identical, return True, else return False. # Edge Cases to Consider * The lengths of the input strings are not the same. * Both input strings are empty.
answer:def are_anagrams(s1, s2): if len(s1) != len(s2): return False char_count = [0] * 26 for char in s1: char_count[ord(char) - ord('a')] += 1 for char in s2: char_count[ord(char) - ord('a')] -= 1 if char_count[ord(char) - ord('a')] < 0: return False return True
question:# Matrix Multiplication Challenge **Context**: You are developing a component for a scientific computing application that requires frequently performing matrix multiplications. Your task is to implement the matrix multiplication operation in a way that validates input dimensions and correctly computes the product of the matrices. **Task**: Write a function `matrix_multiply` that takes two 2D lists (matrices), `multiplicand` and `multiplier`, and returns their matrix product. If the matrices are not compatible for multiplication (i.e., the number of columns in `multiplicand` does not match the number of rows in `multiplier`), the function should raise an appropriate exception. **Function Signature**: ```python def matrix_multiply(multiplicand: list, multiplier: list) -> list: pass ``` **Input**: - `multiplicand` (List[List[int]]): A list of lists representing the first matrix with dimensions m x n. - `multiplier` (List[List[int]]): A list of lists representing the second matrix with dimensions n x p. **Output**: - A list of lists representing the resultant matrix with dimensions m x p. **Constraints**: - Each matrix can have up to 100 rows and columns. - Each element in the matrices is an integer between -1000 and 1000. - Assume that the input matrices are non-empty and contain valid integers. **Example**: ```python multiplicand = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ] multiplier = [ [9, 8, 7], [6, 5, 4], [3, 2, 1] ] result = matrix_multiply(multiplicand, multiplier) # Expected output: # [ # [30, 24, 18], # [84, 69, 54], # [138, 114, 90] # ] ``` **Performance Requirements**: - The function should work efficiently within the provided constraints. - Consider ways to minimize the number of operations in the innermost loop.
answer:def matrix_multiply(multiplicand: list, multiplier: list) -> list: Multiplies two matrices (2D lists) and returns the resultant matrix. Raises ValueError if the matrices cannot be multiplied (i.e., inner dimensions do not match). # Get the dimensions of the matrices m = len(multiplicand) n = len(multiplicand[0]) p = len(multiplier[0]) if n != len(multiplier): raise ValueError("Cannot multiply matrices: Inner dimensions do not match.") # Initialize the result matrix with zeros result = [[0] * p for _ in range(m)] # Perform multiplication for i in range(m): for j in range(p): result[i][j] = sum(multiplicand[i][k] * multiplier[k][j] for k in range(n)) return result