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question:def find_kth_number(k: int) -> int: Returns the k-th number in the defined sequence. >>> find_kth_number(1) 1 >>> find_kth_number(2) 2 >>> find_kth_number(3) 4 >>> find_kth_number(4) 8 >>> find_kth_number(5) 16 >>> find_kth_number(10) 512 >>> find_kth_number(20) 524288 >>> find_kth_number(30) 536870912 >>> find_kth_number(50) 562949953421312 >>> find_kth_number(100) 633825300114114700748351602688 >>> find_kth_number(100000) 2**99999

answer:def find_kth_number(k): Returns the k-th number in the defined sequence. Parameters: k (int): The position of the desired number in the sequence (1-based). Returns: int: The k-th number in the sequence. # The sequence is of the form: 1, 2, 4, 8, 16, ... # Essentially, it's 2^(k-1), because at k=1 the number is 2^(1-1) = 1, # at k=2 the number is 2^(2-1) = 2, at k=3 the number is 2^(3-1) = 4, and so on. return 2**(k-1)

question:def is_valid_chocolate(grid): Function to determine if there exists at least one rectangular sub-grid composed entirely of 'C' in the given grid. Args: grid (list of list of str): A 2D list representing the chocolate grid containing 'C' and 'X'. Returns: str: "YES" if there exists a rectangular sub-grid composed entirely of 'C', otherwise "NO". >>> is_valid_chocolate([ ['C', 'C', 'C', 'C'], ['C', 'X', 'X', 'C'], ['C', 'X', 'X', 'C'], ['C', 'C', 'C', 'C'], ]) 'YES' >>> is_valid_chocolate([ ['C', 'X', 'C'] ]) 'YES' >>> is_valid_chocolate([ ['X', 'X', 'X'], ['X', 'X', 'X'], ]) 'NO' >>> is_valid_chocolate([ ['C', 'X', 'X'], ['X', 'X', 'X'], ['X', 'X', 'C'] ]) 'YES' >>> is_valid_chocolate([ ['X'] ]) 'NO' >>> is_valid_chocolate([ ['C', 'C', 'C'], ['C', 'C', 'C'], ['C', 'C', 'C'] ]) 'YES'

answer:def is_valid_chocolate(grid): Function to determine if there exists at least one rectangular sub-grid composed entirely of 'C' in the given grid. Args: grid (list of list of str): A 2D list representing the chocolate grid containing 'C' and 'X'. Returns: str: "YES" if there exists a rectangular sub-grid composed entirely of 'C', otherwise "NO". n = len(grid) m = len(grid[0]) for i in range(n): for j in range(m): if grid[i][j] == 'C': # Check for 1x1 sub-grid first return "YES" return "NO"

question:from typing import List, Tuple, Union def find_min_cost(k: int, l: int, connections: List[Tuple[int, int, int]]) -> Union[int, str]: Returns the minimal total cost to connect all the villages, or "Impossible" if it's not feasible. >>> find_min_cost(4, 5, [(1, 2, 5), (1, 3, 3), (2, 3, 7), (2, 4, 6), (3, 4, 4)]) 12 >>> find_min_cost(3, 3, [(1, 2, 1), (1, 3, 4), (2, 3, 2)]) 3 >>> find_min_cost(2, 1, [(1, 2, 10)]) 10 >>> find_min_cost(3, 0, []) 'Impossible' >>> find_min_cost(1, 0, []) 0 # Implement the function here def test_example1(): k, l = 4, 5 connections = [ (1, 2, 5), (1, 3, 3), (2, 3, 7), (2, 4, 6), (3, 4, 4) ] assert find_min_cost(k, l, connections) == 12 def test_example2(): k, l = 3, 3 connections = [ (1, 2, 1), (1, 3, 4), (2, 3, 2) ] assert find_min_cost(k, l, connections) == 3 def test_example3(): k, l = 2, 1 connections = [ (1, 2, 10) ] assert find_min_cost(k, l, connections) == 10 def test_example4(): k, l = 3, 0 connections = [] assert find_min_cost(k, l, connections) == "Impossible" def test_single_village(): k, l = 1, 0 connections = [] assert find_min_cost(k, l, connections) == 0 def test_all_villages_connected(): k, l = 4, 6 connections = [ (1, 2, 1), (1, 3, 2), (2, 3, 3), (3, 4, 4), (2, 4, 5), (1, 4, 6) ] assert find_min_cost(k, l, connections) == 7 def test_higher_cost_edge_irrelevant(): k, l = 4, 5 connections = [ (1, 2, 5), (1, 3, 3), (2, 3, 7), (2, 4, 6), (3, 4, 15) ] assert find_min_cost(k, l, connections) == 14

answer:def find_min_cost(k, l, connections): Returns the minimal total cost to connect all the villages, or "Impossible" if it's not feasible. if k == 1: return 0 # Only one village, no cost needed parent = list(range(k+1)) def find(x): if parent[x] != x: parent[x] = find(parent[x]) return parent[x] def union(x, y): rootX = find(x) rootY = find(y) if rootX != rootY: parent[rootX] = rootY connections.sort(key=lambda x: x[2]) total_cost = 0 edges_used = 0 for x, y, c in connections: if find(x) != find(y): union(x, y) total_cost += c edges_used += 1 if edges_used == k - 1: return total_cost return "Impossible" # Usage example: # k, l = 4, 5 # connections = [ # (1, 2, 5), # (1, 3, 3), # (2, 3, 7), # (2, 4, 6), # (3, 4, 4) # ] # print(find_min_cost(k, l, connections)) # Output: 12

question:from typing import List def generate_groups_of_three(n: int, employees: List[str]) -> List[str]: Generates all unique groups of exactly 3 employees and outputs them sorted lexicographically. :param n: int, the number of employees :param employees: list of strings, names of employees :return: list of strings, each string representing a group of 3 employees >>> generate_groups_of_three(4, ["alice", "bob", "carol", "dave"]) ["alice bob carol", "alice bob dave", "alice carol dave", "bob carol dave"] >>> generate_groups_of_three(5, ["eve", "diana", "frank", "george", "harry"]) ["diana eve frank", "diana eve george", "diana eve harry", "diana frank george", "diana frank harry", "diana george harry", "eve frank george", "eve frank harry", "eve george harry", "frank george harry"]

answer:from itertools import combinations def generate_groups_of_three(n, employees): Generates all unique groups of exactly 3 employees and outputs them sorted lexicographically. :param n: int, the number of employees :param employees: list of strings, names of employees :return: list of strings, each string representing a group of 3 employees employees.sort() groups = list(combinations(employees, 3)) return [' '.join(group) for group in groups]