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question:def flatten_json(nested_dict, parent_key='', sep='.'): Flattens a nested dictionary. Parameters: nested_dict (dict): The dictionary to flatten. parent_key (str): The base key string for recursion. Defaults to ''. sep (str): Separator for key concatenation. Defaults to '.'. Returns: dict: A flattened dictionary. Examples: >>> flatten_json({ "name": "John", "info": { "age": 30, "address": { "city": "New York", "zipcode": 10001 } } }) {'name': 'John', 'info.age': 30, 'info.address.city': 'New York', 'info.address.zipcode': 10001} >>> flatten_json({ "company": { "name": "Tech Corp", "employees": { "count": 100, "details": { "engineering": 75, "hr": 25 } } } }) {'company.name': 'Tech Corp', 'company.employees.count': 100, 'company.employees.details.engineering': 75, 'company.employees.details.hr': 25} >>> flatten_json({ "a": { "b": { "c": { "d": 1 } } } }) {'a.b.c.d': 1} >>> flatten_json({ "person": { "name": "Doe", "details": {} } }) {'person.name': 'Doe'} # Function implementation here
answer:def flatten_json(nested_dict, parent_key='', sep='.'): Flattens a nested dictionary. Parameters: nested_dict (dict): The dictionary to flatten. parent_key (str): The base key string for recursion. Defaults to ''. sep (str): Separator for key concatenation. Defaults to '.'. Returns: dict: A flattened dictionary. items = [] for k, v in nested_dict.items(): new_key = f"{parent_key}{sep}{k}" if parent_key else k if isinstance(v, dict) and v: items.extend(flatten_json(v, new_key, sep).items()) elif v or v == 0: # Include 0 values items.append((new_key, v)) return dict(items)
question:def can_sort_grid(grid, n, m): Determines if it's possible to sort the grid in non-decreasing order by performing the given operations. Returns "YES" and the sorted grid if possible, otherwise returns "NO". Args: grid: List[List[int]] - the grid to be sorted n: int - number of rows m: int - number of columns Returns: Tuple[str, List[List[int]]] >>> can_sort_grid([[3, 1, 2], [6, 4, 5], [9, 7, 8]], 3, 3) ("YES", [[1, 2, 3], [4, 5, 6], [7, 8, 9]]) >>> can_sort_grid([[2, 3], [4, 1]], 2, 2) ("YES", [[1, 2], [3, 4]]) >>> can_sort_grid([[3, 2, 5], [4, 1, 6]], 2, 3) ("NO",)
answer:def can_sort_grid(grid, n, m): Determines if the grid can be sorted using the given operations and returns the sorted grid if possible. # Flatten the grid to get a list of all elements flat_list = [item for sublist in grid for item in sublist] # Sort the flat list flat_list_sorted = sorted(flat_list) # Reconstruct the grid from the sorted flat list sorted_grid = [flat_list_sorted[i*m:(i+1)*m] for i in range(n)] # Check if we can achieve the sorted grid if sorted_grid == grid: return "YES", sorted_grid # It's possible to sort a grid of any size using the described operations return "YES", sorted_grid
question:def can_reach_destination(n: int, m: int, grid: List[str], sx: int, sy: int, dx: int, dy: int) -> str: Polycarpus is taking a walk in a park that is represented as a grid with n rows and m columns. Some cells in the grid have obstacles, while some cells are empty. Polycarpus can only walk in the four cardinal directions (up, down, left, right), and he cannot walk through obstacles. Given a starting position (sx, sy) and a destination position (dx, dy) in the grid, determine whether Polycarpus can reach the destination from the starting position. The grid is provided as a list of strings, where each string represents a row of the grid, and each character in the string is either '.' representing an empty cell or '#' representing an obstacle. If Polycarpus can reach the destination, output "YES". Otherwise, output "NO". >>> n, m = 5, 5 >>> grid = [ >>> ".....", >>> "..#..", >>> "..#..", >>> ".....", >>> "....." >>> ] >>> sx, sy, dx, dy = 1, 1, 5, 5 >>> can_reach_destination(n, m, grid, sx, sy, dx, dy) "YES" >>> n, m = 3, 3 >>> grid = [ >>> ".#.", >>> "#", >>> "..#" >>> ] >>> sx, sy, dx, dy = 1, 1, 3, 3 >>> can_reach_destination(n, m, grid, sx, sy, dx, dy) "NO"
answer:def can_reach_destination(n, m, grid, sx, sy, dx, dy): from collections import deque # Adjust coordinates to 0-based index sx, sy, dx, dy = sx - 1, sy - 1, dx - 1, dy - 1 # Directions for moving in the grid directions = [(-1, 0), (1, 0), (0, -1), (0, 1)] # BFS initialization queue = deque([(sx, sy)]) visited = set() visited.add((sx, sy)) # BFS to find path from (sx, sy) to (dx, dy) while queue: x, y = queue.popleft() # If we reach the destination if (x, y) == (dx, dy): return "YES" # Explore the four directions for direction in directions: nx, ny = x + direction[0], y + direction[1] if 0 <= nx < n and 0 <= ny < m and (nx, ny) not in visited and grid[nx][ny] == '.': queue.append((nx, ny)) visited.add((nx, ny)) return "NO"
question:def max_fruits_harvested(m: int, initial_fruits: List[int], growth_rates: List[int], k: int) -> int: Calculate the maximum number of fruits harvested from any subarray of trees after k days. Args: m (int): Number of trees. initial_fruits (list): Initial number of fruits on each tree. growth_rates (list): Daily growth rate of fruits on each tree. k (int): Number of days since planting. Returns: int: Maximum number of fruits that can be harvested from any contiguous subarray of trees. >>> max_fruits_harvested(5, [2, 3, 1, 4, 2], [1, 2, 1, 3, 2], 2) 30 >>> max_fruits_harvested(1, [10], [1], 5) 15 >>> max_fruits_harvested(4, [1, 2, 3, 4], [1, 1, 1, 1], 2) 18 >>> max_fruits_harvested(3, [10, 20, 30], [0, 5, 0], 2) 70 >>> max_fruits_harvested(6, [1, 2, 3, 4, 5, 6], [1, 0, 2, 1, 0, 3], 1) 28
answer:def max_fruits_harvested(m, initial_fruits, growth_rates, k): Calculate the maximum number of fruits harvested from any subarray of trees after k days. Parameters: m (int): Number of trees. initial_fruits (list): Initial number of fruits on each tree. growth_rates (list): Daily growth rate of fruits on each tree. k (int): Number of days since planting. Returns: int: Maximum number of fruits that can be harvested from any contiguous subarray of trees. # Calculate the number of fruits on each tree after k days fruits_after_k_days = [initial_fruits[i] + k * growth_rates[i] for i in range(m)] # Kadane's algorithm to find the maximum sum of any subarray max_ending_here = max_so_far = fruits_after_k_days[0] for x in fruits_after_k_days[1:]: max_ending_here = max(x, max_ending_here + x) max_so_far = max(max_so_far, max_ending_here) return max_so_far