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question:Given the class definitions for a `Node`, `DirectedEdge`, and a `DirectedGraph`, implement a function that performs a Depth-First Search (DFS) starting from a given node in a directed graph. Your task is to determine if there is a path from the starting node to all other nodes in the graph. # Function Signature: ```python def dfs_traversal(graph: DirectedGraph, start_node: str) -> List[str]: Perform DFS on the given directed graph starting from the start_node. Parameters: - graph: DirectedGraph instance representing the directed graph. - start_node: A string representing the name of the starting node. Returns: - List of node names in the order they are visited during the DFS traversal, starting from start_node. ``` # Input: - A `DirectedGraph` object which already contains the graph nodes and edges. - A string representing the name of the starting node. # Output: - A list of node names in the order they were visited during DFS traversal starting from the start_node. # Constraints: - Assume all node names are unique. - The starting node will always be part of the graph. - The graph may contain cycles and disconnected components. # Example: ```python # Constructing a sample graph graph_dict = { 'A': ['B', 'C'], 'B': ['D', 'E'], 'C': ['F'], 'D': [], 'E': ['F'], 'F': [] } graph = DirectedGraph(graph_dict) # Running the dfs_traversal function print(dfs_traversal(graph, 'A')) # Output: ['A', 'B', 'D', 'E', 'F', 'C'] ``` # Note: - The order of node visits may vary as long as it follows the DFS traversal method. Multiple correct answers may exist due to the nature of graph traversal. # Additional Requirements: - Ensure the function handles graphs with cycles without infinite recursion. - The function should be able to handle disconnected nodes gracefully.
answer:class Node: def __init__(self, name: str): self.name = name self.edges = [] def add_edge(self, node): self.edges.append(node) class DirectedGraph: def __init__(self): self.nodes = {} def add_node(self, name: str): if name not in self.nodes: self.nodes[name] = Node(name) def add_edge(self, from_node: str, to_node: str): if from_node in self.nodes and to_node in self.nodes: self.nodes[from_node].add_edge(self.nodes[to_node]) def dfs_traversal(graph: DirectedGraph, start_node: str) -> list: visited = set() result = [] def dfs(node): if node.name not in visited: visited.add(node.name) result.append(node.name) for edge in node.edges: dfs(edge) start = graph.nodes[start_node] dfs(start) return result
question:You are provided with a function for generating prime numbers less than a given integer ( n ) using a modified version of the Sieve of Eratosthenes algorithm. Your task is to extend this functionality in the following ways: 1. **Prime Count**: Write a function `count_primes(n)` which uses `get_primes(n)` and returns the count of prime numbers less than ( n ). 2. **Prime Sum**: Write a function `sum_primes(n)` which uses `get_primes(n)` and returns the sum of all prime numbers less than ( n ). # Expected Input and Output * **Input**: * `n` (integer): A positive integer greater than or equal to 2. * **Output**: * `count_primes(n)`: An integer representing the number of prime numbers less than ( n ). * `sum_primes(n)`: An integer representing the sum of all prime numbers less than ( n ). # Function Signatures ```python def count_primes(n: int) -> int: pass def sum_primes(n: int) -> int: pass ``` # Examples ```python # Example 1 n = 10 print(count_primes(n)) # Output: 4 print(sum_primes(n)) # Output: 17 # Example 2 n = 20 print(count_primes(n)) # Output: 8 print(sum_primes(n)) # Output: 77 ``` # Constraints * ( n ) will be a positive integer greater than or equal to 2. * Ensure that your implementations of `count_primes` and `sum_primes` are optimized for performance.
answer:def get_primes(n): Return a list of all prime numbers less than n if n <= 2: return [] sieve = [True] * n sieve[0] = sieve[1] = False for start in range(2, int(n**0.5) + 1): if sieve[start]: for i in range(start*start, n, start): sieve[i] = False return [num for num, is_prime in enumerate(sieve) if is_prime] def count_primes(n): Returns the count of prime numbers less than n primes = get_primes(n) return len(primes) def sum_primes(n): Returns the sum of all prime numbers less than n primes = get_primes(n) return sum(primes)
question:You are tasked with writing a function that computes the minimum number of bit flips required to convert one integer to another. Given two integers, `A` and `B`, determine the number of differing bits between their binary representations. # Details * **Function Name**: `minimal_bit_flips` * **Input**: Two integers, `A` and `B`. * **Output**: An integer representing the minimal number of bit flips required to transform `A` into `B`. # Input & Output Format * `A` and `B` will be non-negative integers (0 ≤ `A`, `B` ≤ 10^9). **Example**: ```python assert minimal_bit_flips(29, 15) == 2 assert minimal_bit_flips(7, 8) == 4 assert minimal_bit_flips(123, 456) == 6 ``` # Constraints * Your solution should handle edge cases efficiently. * Performance is critical; the function should operate within an acceptable time frame for the given constraints. # Explanation: 1. Compute the XOR of `A` and `B`. 2. Count the number of `1`s in the resulting binary number - each `1` represents a differing bit. 3. Return the count of these `1`s. # Function Signature ```python def minimal_bit_flips(A: int, B: int) -> int: # Implementation here ``` Developers should ensure they handle large inputs efficiently and that their solution is optimized for performance within the provided constraints.
answer:def minimal_bit_flips(A, B): Returns the minimal number of bit flips required to convert integer A to integer B. # Compute XOR of A and B x = A ^ B # Count the number of 1s in the binary representation of the XOR result count = bin(x).count('1') return count
question:# Question: Implement an Optimized Power Function with Detailed Analysis You're required to implement a power function that calculates `a` raised to the power `n` efficiently. The function should also be capable of returning the result modulo `mod` if `mod` is specified. Implement it using both an iterative and a recursive approach to demonstrate your understanding of different problem-solving methods. Function Signature ```python def power(a: int, n: int, mod: int = None) -> int: pass def power_recur(a: int, n: int, mod: int = None) -> int: pass ``` # Input * An integer `a` (1 <= |a| <= 10^9) representing the base. * An integer `n` (0 <= n <= 10^9) representing the exponent. * An optional integer `mod` (2 <= mod <= 10^9), which is the number to compute the result modulo. If `mod` is not provided, you should return the result as a standard integer. # Output * The function should return the result of `a` raised to the power `n`. If `mod` is given, return `a^n % mod`. # Constraints * Ensure that your implementations are efficient to avoid timeouts for large inputs. * Handle edge cases such as `n` being zero, `a` being one, and handle the optional `mod` correctly. # Example ```python # Examples without mod print(power(2, 10)) # Output: 1024 print(power_recur(2, 10)) # Output: 1024 # Examples with mod print(power(2, 10, 1000)) # Output: 24 print(power_recur(2, 10, 1000)) # Output: 24 # Edge cases print(power(2, 0)) # Output: 1 print(power_recur(2, 0)) # Output: 1 ``` # Explanation 1. **No Modulo**: For `power(2, 10)`, the function should return `2^10 = 1024`. 2. **With Modulo**: For `power(2, 10, 1000)`, the function should return `(2^10) % 1000 = 24`. 3. **Edge Cases**: For `power(2, 0)`, the function should return `1` since any number to the power of zero is one. Ensure to consider efficient implementation principles during coding to handle large numbers and prevent overflow errors where applicable.
answer:def power(a: int, n: int, mod: int = None) -> int: Iterative method to calculate a raised to the power n efficiently. If mod is provided, return the result modulo mod. result = 1 base = a if mod is None else a % mod while n > 0: if n % 2 == 1: result = result * base if mod is None else (result * base) % mod base = base * base if mod is None else (base * base) % mod n //= 2 return result def power_recur(a: int, n: int, mod: int = None) -> int: Recursive method to calculate a raised to the power n efficiently. If mod is provided, return the result modulo mod. if n == 0: return 1 half_power = power_recur(a, n // 2, mod) if mod is not None: half_power = half_power % mod if n % 2 == 0: return (half_power * half_power) % mod if mod is not None else half_power * half_power else: return (half_power * half_power * a) % mod if mod is not None else half_power * half_power * a