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question:def min_path_cost(grid): Returns the minimum cost to traverse from the top-left to the bottom-right corner of the grid. >>> min_path_cost([[1, 3, 1], [1, 5, 1], [4, 2, 1]]) 7 >>> min_path_cost([[5]]) 5 >>> min_path_cost([[1, 2, 3, 4]]) 10 >>> min_path_cost([[1], [2], [3], [4]]) 10 >>> min_path_cost([]) 0
answer:def min_path_cost(grid): Returns the minimum cost to traverse from the top-left to the bottom-right corner of the grid. if not grid or not grid[0]: return 0 m, n = len(grid), len(grid[0]) dp = [[0] * n for _ in range(m)] dp[0][0] = grid[0][0] for i in range(1, m): dp[i][0] = dp[i-1][0] + grid[i][0] for j in range(1, n): dp[0][j] = dp[0][j-1] + grid[0][j] for i in range(1, m): for j in range(1, n): dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j] return dp[m-1][n-1]
question:def count_top_scorers(scores: List[int]) -> int: Returns the number of players who have the highest score. Args: scores (List[int]): A list of integers representing the scores of players. Returns: int: The number of players who have the highest score. Examples: >>> count_top_scorers([85, 90, 78, 90, 88]) 2 >>> count_top_scorers([77, 68, 55, 77, 70, 77]) 3 >>> count_top_scorers([50, 65, 80, 99, 45, 99, 78]) 2
answer:def count_top_scorers(scores): Returns the number of players who have the highest score. Args: scores (List[int]): A list of integers representing the scores of players. Returns: int: The number of players who have the highest score. if not scores: return 0 highest_score = max(scores) return scores.count(highest_score)
question:def organize_books(N, borrowed_books): Organizes books in the order they have been borrowed and alphabetically if borrowing counts are the same. Parameters: N (int): Number of borrowed books borrowed_books (list): List containing names of borrowed books Returns: tuple: A tuple where the first element is the number of distinct books, and the second element is a list with book names sorted by borrow count and then alphabetically. Examples: >>> organize_books(10, ["book a", "book b", "book c", "book b", "book d", "book a", "book c", "book b", "book e", "book e"]) (5, ["book d", "book a", "book c", "book e", "book b"]) >>> organize_books(5, ["the great gatsby", "a tale of two cities", "war and peace", "a tale of two cities", "the great gatsby"]) (3, ["war and peace", "a tale of two cities", "the great gatsby"]) >>> organize_books(4, ["book f", "book d", "book e", "book c"]) (4, ["book c", "book d", "book e", "book f"]) >>> organize_books(5, ["book a", "book a", "book a", "book a", "book a"]) (1, ["book a"]) >>> organize_books(3, ["alpha", "beta", "gamma"]) (3, ["alpha", "beta", "gamma"]) >>> organize_books(1, ["onlybook"]) (1, ["onlybook"])
answer:def organize_books(N, borrowed_books): Organizes books in the order they have been borrowed and alphabetically if borrowing counts are the same. Parameters: N (int): Number of borrowed books borrowed_books (list): List containing names of borrowed books Returns: tuple: A tuple where the first element is the number of distinct books, and the second element is a list with book names sorted by borrow count and then alphabetically. from collections import Counter # Count the occurrences of each book book_count = Counter(borrowed_books) # Sort books first by count (ascending) and then by name (alphabetically) sorted_books = sorted(book_count.items(), key=lambda x: (x[1], x[0])) # Extract the sorted book names sorted_book_names = [book[0] for book in sorted_books] return (len(book_count), sorted_book_names)
question:def process_queries(n: int, q: int, b: List[int], queries: List[List[int]]) -> List[int]: You are given a sequence b_1, b_2, ..., b_n. The task is to answer the following types of queries on it: Type 1. Given three integers l, r, and x (1 <= l <= r <= n), add the integer x to every element in the subarray [l, r]. Type 2. Given two integers l and r (1 <= l <= r <= n), find the maximum value in the subarray [l, r]. Args: n (int): Length of the sequence. q (int): The number of queries. b (List[int]): Initial sequence. queries (List[List[int]]): List of queries. Returns: List[int]: List of results for each Type 2 query. >>> process_queries(6, 5, [3, 1, 4, 1, 5, 9], [[1, 2, 4, 2], [2, 1, 3], [1, 3, 6, 1], [2, 2, 6], [2, 4, 4]]) [6, 10, 4] >>> process_queries(3, 2, [5, 6, 7], [[2, 1, 2], [2, 1, 3]]) [6, 7] >>> process_queries(4, 3, [1, 2, 3, 4], [[1, 1, 4, 1], [1, 2, 3, 2], [2, 1, 3]]) [6] >>> process_queries(5, 4, [1, 2, 3, 4, 5], [[1, 3, 3, 10], [2, 3, 3], [1, 1, 5, 5], [2, 4, 5]]) [13, 10]
answer:def process_queries(n, q, b, queries): results = [] for query in queries: if query[0] == 1: l, r, x = query[1] - 1, query[2] - 1, query[3] for i in range(l, r + 1): b[i] += x elif query[0] == 2: l, r = query[1] - 1, query[2] - 1 max_value = max(b[l:r + 1]) results.append(max_value) return results