answer:def calculate_similarity_scores(category_descriptions, keywords, queries): def keyword_frequencies(description): words = description.split() freq = {keyword: 0 for keyword in keywords} for word in words: if word in freq: freq[word] += 1 return freq frequencies = [keyword_frequencies(desc) for desc in category_descriptions] def similarity_score(freq1, freq2): score = 0 for keyword in keywords: score += abs(freq1[keyword] - freq2[keyword]) return score results = [] for a, b in queries: freq1 = frequencies[a-1] freq2 = frequencies[b-1] results.append(similarity_score(freq1, freq2)) return results

answer:def calculate_distances(test_cases): Given a list of test cases (each a tuple of four integers representing two points on a grid), returns a list of distances between these points using the formula |x1 - x2| + |y1 - y2|. distances = [] for x1, y1, x2, y2 in test_cases: distance = abs(x1 - x2) + abs(y1 - y2) distances.append(distance) return distances def parse_input(input_str): Parses an input string in the format specified in the problem description and returns the number of test cases and a list of test cases. lines = input_str.strip().split('n') T = int(lines[0].strip()) test_cases = [] for i in range(1, T + 1): x1, y1, x2, y2 = map(int, lines[i].strip().split()) test_cases.append((x1, y1, x2, y2)) return T, test_cases def format_output(distances): Formats a list of distances (integers) into an output string with each distance on a new line. return 'n'.join(map(str, distances)) def main(input_str): T, test_cases = parse_input(input_str) distances = calculate_distances(test_cases) return format_output(distances)

answer:def find_minimum_repair_cost(n, m, roads): Finds the minimum total length of roads that need repair to keep all intersections connected. Parameters: n (int): Number of intersections. m (int): Number of roads. roads (List[Tuple[int, int, int]]): List of tuples where each tuple represents (u, v, w). Returns: int: Minimum total length of roads that need repair. # Using Kruskal's algorithm to find the Minimum Spanning Tree (MST) parent = list(range(n + 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[rootY] = rootX # Sort the roads by their weight (length) roads.sort(key=lambda x: x[2]) mst_cost = 0 edges_used = 0 for u, v, w in roads: if find(u) != find(v): union(u, v) mst_cost += w edges_used += 1 # Early stop if we've used n-1 edges (which is enough to connect all intersections) if edges_used == n - 1: break return mst_cost

answer:import string def is_pangram(sentence): Determines if a given sentence is a pangram. Ignores spaces and punctuation, case insensitive. alphabet_set = set(string.ascii_lowercase) sentence_set = set(c.lower() for c in sentence if c.isalpha()) return "YES" if alphabet_set <= sentence_set else "NO"