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question:# Binary Search Tree with Height Calculation Objective Expand the given Binary Search Tree (BST) implementation by adding a method to calculate the height of the tree. Demonstrate your understanding of recursively determining the height of a tree structure. Problem Statement A Binary Search Tree (BST) class is provided with methods to insert nodes, search for a value, calculate size, and traverse nodes in different orders. Your task is to implement an additional method in the `BST` class to compute the height of the tree. Definitions - **Height of a Tree**: The height of a tree is the number of edges on the longest path from the root to a leaf. Function Signature ```python class BST(object): ... def height(self): pass ``` Expected Input and Output - **Input**: None, the height method operates on the BST instance. - **Output**: An integer representing the height of the BST. Constraints - The tree can have `N` nodes where `0 <= N <= 10^4`. - Only positive integers are stored in the tree nodes. Examples Consider the same tree structure created previously: ``` 10 / 6 15 / / 4 9 12 24 / / 7 20 30 / 18 ``` - **Input**: No input required, call `tree.height()` - **Output**: `4` Performance Requirements - Aim to maintain the overall time complexity of checking the height to O(N) where `N` is the number of nodes. - Ensure optimized space usage by managing recursion depth appropriately. Steps to Solve 1. Add a new method named `height` to the `BST` class. 2. Implement the height calculation by recursively determining the height of each subtree. 3. Ensure the method handles edge cases such as an empty tree correctly.

answer:class Node: def __init__(self, key): self.left = None self.right = None self.val = key class BST: def __init__(self): self.root = None def insert(self, key): if self.root is None: self.root = Node(key) else: self._insert(self.root, key) def _insert(self, root, key): if key < root.val: if root.left is None: root.left = Node(key) else: self._insert(root.left, key) else: if root.right is None: root.right = Node(key) else: self._insert(root.right, key) def height(self): return self._height(self.root) def _height(self, node): if node is None: return -1 left_height = self._height(node.left) right_height = self._height(node.right) return max(left_height, right_height) + 1

question:# Coding Challenge: Efficient Range Queries with Segment Tree Problem Statement You are tasked to implement a `SegmentTree` class that supports the following operations: 1. Build a segment tree from an initial array using a given commutative function. 2. Perform range queries on the array using the segment tree. 3. Update elements in the array and reflect the change in the segment tree. Specifications Implement the `SegmentTree` class with the following methods: 1. `__init__(self, arr, function)`: - Initializes the Segment Tree with the array `arr` and the commutative function `function`. 2. `update(self, p, v)`: - Updates the element at index `p` in the array to value `v` and updates the segment tree accordingly. 3. `query(self, l, r)`: - Returns the result of the commutative function applied to the range `[l, r]` in the array. Function Signatures ```python class SegmentTree: def __init__(self, arr, function): # Your code here def update(self, p, v): # Your code here def query(self, l, r): # Your code here ``` Example ```python # Example 1: Using max function mytree = SegmentTree([2, 4, 5, 3, 4], max) print(mytree.query(2, 4)) # Output: 5 mytree.update(3, 6) print(mytree.query(0, 3)) # Output: 6 # Example 2: Using sum (lambda function) mytree = SegmentTree([4, 5, 2, 3, 4, 43, 3], lambda a, b: a + b) print(mytree.query(0, 6)) # Output: 64 mytree.update(2, -10) print(mytree.query(0, 6)) # Output: 52 # Example 3: Using tuple addition (lambda function) mytree = SegmentTree([(1, 2), (4, 6), (4, 5)], lambda a, b: (a[0] + b[0], a[1] + b[1])) print(mytree.query(0, 2)) # Output: (9, 13) mytree.update(2, (-1, 2)) print(mytree.query(0, 2)) # Output: (4, 10) ``` Constraints 1. The array length (N) will be in the range [1, 10^5]. 2. The commutative function will be applicable to the array elements. 3. The number of queries and updates (Q) will be in the range [1, 10^5]. **Note**: For this problem, assume all inputs are correctly formatted and valid as per the constraints.

answer:class SegmentTree: def __init__(self, arr, function): self.n = len(arr) self.function = function self.tree = [0] * (2 * self.n) self.build(arr) def build(self, arr): # Initialize leaves in the segment tree for i in range(self.n): self.tree[self.n + i] = arr[i] # Initialize internal nodes for i in range(self.n - 1, 0, -1): self.tree[i] = self.function(self.tree[i * 2], self.tree[i * 2 + 1]) def update(self, p, v): # Update the value at index p p += self.n self.tree[p] = v # Update internal nodes while p > 1: p //= 2 self.tree[p] = self.function(self.tree[p * 2], self.tree[p * 2 + 1]) def query(self, l, r): # Perform range query l += self.n r += self.n res = None while l <= r: if l % 2 == 1: res = self.tree[l] if res is None else self.function(res, self.tree[l]) l += 1 if r % 2 == 0: res = self.tree[r] if res is None else self.function(res, self.tree[r]) r -= 1 l //= 2 r //= 2 return res

question:# Graph Coloring Challenge Problem Statement: Given an undirected graph represented by an adjacency matrix, determine whether the graph is bipartite. A graph is bipartite if its vertices can be divided into two independent sets (U) and (V) such that every edge (e) connects a vertex in (U) to one in (V). Input: - A 2D list `adj_matrix` where `adj_matrix[i][j]` is 1 if there is an edge between vertex `i` and vertex `j`, and 0 otherwise. It is guaranteed that `adj_matrix[i][i]` will be 0 (no self-loops). Output: - Return `True` if the graph is bipartite, `False` otherwise. Constraints: - (1 leq N leq 1000) where (N) is the number of vertices in the graph. Example: ```python # Example 1: adj_matrix = [ [0, 1, 0, 1], [1, 0, 1, 0], [0, 1, 0, 1], [1, 0, 1, 0] ] print(is_bipartite(adj_matrix)) # Output: True # Example 2: adj_matrix = [ [0, 1, 1], [1, 0, 1], [1, 1, 0] ] print(is_bipartite(adj_matrix)) # Output: False ``` Implementation: Write a function `is_bipartite(adj_matrix)` that takes the adjacency matrix as input and returns whether the graph is bipartite. # Function Signature: ```python def is_bipartite(adj_matrix: list[list[int]]) -> bool: pass ```

answer:from collections import deque def is_bipartite(adj_matrix): Determines if a given graph is bipartite. Parameters: adj_matrix (list of list of int): The adjacency matrix of the graph. Returns: bool: True if the graph is bipartite, False otherwise. N = len(adj_matrix) colors = [-1] * N # -1 means uncolored, 0 and 1 are the two different colors def bfs(start): queue = deque([start]) colors[start] = 0 # Start coloring with color 0 while queue: node = queue.popleft() for neighbor in range(N): if adj_matrix[node][neighbor] == 1: # There is an edge if colors[neighbor] == -1: # If not yet colored colors[neighbor] = 1 - colors[node] # Assign the opposite color queue.append(neighbor) elif colors[neighbor] == colors[node]: # If neighbor has the same color return False return True for i in range(N): if colors[i] == -1: # If the node is not yet colored if not bfs(i): return False return True

question:**Objective**: Demonstrate your understanding of bit manipulation and algorithm implementation with bitwise operations. # Problem Statement You are tasked with writing a function that swaps the odd and even bits of a given non-negative integer. Specifically, bit-0 (the least significant bit) should be swapped with bit-1, bit-2 with bit-3, and so on. # Function Signature ```python def swap_odd_even_bits(n: int) -> int: pass ``` # Input - `n` (int): A non-negative integer (0 ≤ n ≤ 2^31-1). # Output - (int): The integer result after swapping the odd and even bits. # Example - Example 1: ```python swap_odd_even_bits(22) ``` Output: ```plaintext 41 ``` Explanation: - 22 in binary is `010110`. - Swapping odd and even bits: `101001`, which is 41 in decimal. - Example 2: ```python swap_odd_even_bits(10) ``` Output: ```plaintext 5 ``` Explanation: - 10 in binary is `1010`. - Swapping odd and even bits: `0101`, which is 5 in decimal. # Constraints - You should use as few instructions as possible. - You should not use any library functions for the conversion between binary and decimal. - The solution should be efficient in both time and space complexity. # Scenario Consider you are writing software where this functionality is required as part of a low-level network protocol that tags each packet with a uniquely generated bit-swapped identifier for differentiation purposes. Provide a function implementation that meets the above requirements and efficiently handles the swapping of odd and even bits.

answer:def swap_odd_even_bits(n: int) -> int: Swaps the odd and even bits of a given non-negative integer. :param n: A non-negative integer :return: The integer result after swapping the odd and even bits # Odd mask (0xAAAAAAAA) has bits set on odd positions: 10101010... in binary odd_mask = 0xAAAAAAAA # Even mask (0x55555555) has bits set on even positions: 01010101... in binary even_mask = 0x55555555 # Isolate odd bits and shift them to even positions odd_bits_shifted = (n & odd_mask) >> 1 # Isolate even bits and shift them to odd positions even_bits_shifted = (n & even_mask) << 1 # Combine the shifted results return odd_bits_shifted | even_bits_shifted