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question:You need to implement a function, `longest_common_subsequence`, that finds the length of the longest common subsequence between two strings. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, "abc", "abg", "bdf" are subsequences of "abcdefg". **Function Signature:** ```python def longest_common_subsequence(str1: str, str2: str) -> int: ``` **Parameters:** - `str1` (`str`): The first input string. - `str2` (`str`): The second input string. **Returns:** - `int`: The length of the longest common subsequence between the two input strings. **Constraints:** - Both input strings consist of only lowercase English letters. - The lengths of both strings are at most 1000 characters. **Example:** ```python str1 = "abcde" str2 = "ace" print(longest_common_subsequence(str1, str2)) # Output: 3, as the longest common subsequence is "ace" str1 = "abcdxyz" str2 = "xyzabcd" print(longest_common_subsequence(str1, str2)) # Output: 4, as the longest common subsequence is "abcd" ``` **Requirements:** - You should use dynamic programming to solve the problem and ensure the solution runs efficiently within the constraints. - Consider edge cases such as empty strings or no common subsequence between the input strings.
answer:def longest_common_subsequence(str1: str, str2: str) -> int: Finds the length of the longest common subsequence between two strings. m, n = len(str1), len(str2) # Create a (m+1) x (n+1) matrix initialized with zeros dp = [[0] * (n + 1) for _ in range(m + 1)] # Build the matrix from bottom up for i in range(1, m + 1): for j in range(1, n + 1): if str1[i - 1] == str2[j - 1]: dp[i][j] = dp[i - 1][j - 1] + 1 else: dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) return dp[m][n]
question:In this problem, you will construct a function to generate a Fibonacci sequence using recursion, memoization, and bottom-up dynamic programming approaches. The primary goal is to enable a comparative analysis of these three methods in terms of performance and implementation complexity. Your task is to implement the function `fibonacci(n, method)`, where `n` is the position in the Fibonacci sequence, and `method` specifies which of the three methods to use to compute the Fibonacci number. # Function Signature ```python def fibonacci(n, method): Parameters ---------- n : int The position in the Fibonacci sequence (0-indexed). method : str The method to compute the Fibonacci number: 'recursive', 'memoization', or 'bottom-up'. Returns ------- int The Fibonacci number at position n. pass ``` # Requirements 1. If the `method` is `'recursive'`, use a simple recursive function to calculate the Fibonacci number. 2. If the `method` is `'memoization'`, use a recursive function with memoization to store intermediate results. 3. If the `method` is `'bottom-up'`, use an iterative approach to calculate the Fibonacci number using bottom-up dynamic programming. # Example Given the inputs: ```python n = 10 method = 'recursive' ``` The function should return: ```python 55 ``` Another example: ```python n = 10 method = 'memoization' ``` The function should also return: ```python 55 ``` And another example: ```python n = 10 method = 'bottom-up' ``` The function should again return: ```python 55 ``` # Constraints - `n` will be a non-negative integer. - The maximum value for `n` is 30 for the 'recursive' method due to performance considerations. - The 'memoization' and 'bottom-up' methods should handle `n` up to 5000 efficiently.
answer:def fibonacci(n, method): Parameters ---------- n : int The position in the Fibonacci sequence (0-indexed). method : str The method to compute the Fibonacci number: 'recursive', 'memoization', or 'bottom-up'. Returns ------- int The Fibonacci number at position n. if method == 'recursive': return recursive_fib(n) elif method == 'memoization': memo = {} return memoization_fib(n, memo) elif method == 'bottom-up': return bottom_up_fib(n) else: raise ValueError("method should be 'recursive', 'memoization', or 'bottom-up'") def recursive_fib(n): if n <= 1: return n else: return recursive_fib(n - 1) + recursive_fib(n - 2) def memoization_fib(n, memo): if n in memo: return memo[n] if n <= 1: memo[n] = n else: memo[n] = memoization_fib(n - 1, memo) + memoization_fib(n - 2, memo) return memo[n] def bottom_up_fib(n): if n <= 1: return n fib_array = [0] * (n + 1) fib_array[1] = 1 for i in range(2, n + 1): fib_array[i] = fib_array[i - 1] + fib_array[i - 2] return fib_array[n]
question:Consider a list of integers representing the distinct nodes in an undirected graph and a list of tuples representing the undirected edges of the graph, where each tuple contains two integers. For example: ```python nodes = [1, 2, 3, 4] edges = [(1, 2), (2, 3), (3, 4), (4, 1), (1, 3)] ``` Your task is to implement the function `is_cyclic(nodes, edges)` that determines if the graph contains any cycles. The function should return `True` if a cycle is detected, and `False` otherwise. You should use Depth-First Search (DFS) to detect cycles. You are required to use the following components from relevant Python libraries: - `collections.defaultdict` The function signature is: ```python def is_cyclic(nodes: list, edges: list) -> bool: pass ```
answer:from collections import defaultdict def is_cyclic(nodes, edges): Determines if the graph contains any cycles using DFS. Args: nodes: List of integers representing the distinct nodes in the graph. edges: List of tuples, where each tuple contains two integers representing an undirected edge. Returns: bool: True if a cycle is detected, False otherwise. def dfs(v, parent): visited[v] = True for neighbor in graph[v]: if not visited[neighbor]: if dfs(neighbor, v): return True elif parent != neighbor: return True return False graph = defaultdict(list) for edge in edges: graph[edge[0]].append(edge[1]) graph[edge[1]].append(edge[0]) visited = {node: False for node in nodes} for node in nodes: if not visited[node]: if dfs(node, -1): return True return False
question:You are given a partially implemented function named `matrix_multiplication` that performs matrix multiplication on two 2-D NumPy arrays. Matrix multiplication is defined such that if you have two matrices `A` and `B`, their product `C = A * B` is a matrix whose elements are computed as the dot product of the rows of `A` with the columns of `B`. Your task is to complete the `matrix_multiplication` function so that it returns the resulting matrix from multiplying the input matrices `A` and `B`. # Function Signature ```python def matrix_multiplication(A, B): # Check if the number of columns in A is equal to the number of rows in B if A.shape[1] != B.shape[0]: raise ValueError("Number of columns of A must equal number of rows of B.") # Initialize a result matrix with zeros result = np.zeros((A.shape[0], B.shape[1])) # Compute the matrix multiplication for i in range(A.shape[0]): for j in range(B.shape[1]): result[i, j] = np.dot(A[i, :], B[:, j]) return result ``` # Input: - `A` : A 2-D NumPy array with shape `(m, n)`. - `B` : A 2-D NumPy array with shape `(n, p)`. # Output: - `result` : A 2-D NumPy array with shape `(m, p)` containing the result of the matrix multiplication. # Requirements: - Use NumPy's `dot` function to compute the dot product of corresponding rows and columns. # Example: ```python import numpy as np A = np.array([[1, 2], [3, 4]]) B = np.array([[5, 6], [7, 8]]) result = matrix_multiplication(A, B) print("Result:") print(result) # Expected Output # Result: # [[19 22] # [43 50]] ``` # Explanation: 1. **Shape Check:** Ensure the number of columns in matrix `A` is equal to the number of rows in matrix `B`. 2. **Initiate Result Matrix:** Create an output matrix filled with zeros of the appropriate shape. 3. **Element Calculation:** Use nested loops to fill in the result matrix by calculating the dot product of corresponding elements. 4. **Return Result:** Return the result matrix containing the product of `A` and `B`. Your task is to complete the `matrix_multiplication` function as described above, ensuring the correct use of NumPy functions for the calculations.
answer:import numpy as np def matrix_multiplication(A, B): Multiplies two matrices A and B using matrix multiplication rules. Parameters: A (np.ndarray): A 2-D numpy array of shape (m, n). B (np.ndarray): A 2-D numpy array of shape (n, p). Returns: np.ndarray: The resulting matrix of shape (m, p). # Check if the number of columns in A is equal to the number of rows in B if A.shape[1] != B.shape[0]: raise ValueError("Number of columns of A must equal number of rows of B.") # Initialize a result matrix with zeros result = np.zeros((A.shape[0], B.shape[1])) # Compute the matrix multiplication for i in range(A.shape[0]): for j in range(B.shape[1]): result[i, j] = np.dot(A[i, :], B[:, j]) return result