answer:def minimum_sum_after_split_and_reverse(T, test_cases): Returns the minimum sum of the elements of array A after performing the specified split and reverse operations once per each test case. Parameters: T (int): Number of test cases test_cases (list): List containing tuples, each containing an integer N and a list A of N integers Returns: list: List of integers representing the minimum sum of elements for each test case results = [] for test_case in test_cases: N, A = test_case # Since the problem description guarantees the array A and the split of A has no impact on the sum, # Just summarize the array, as sum remains constant. total_sum = sum(A) results.append(total_sum) return results

answer:import heapq def dijkstra(graph, start, end, n): Find the shortest path from start to end in a graph using Dijkstra's algorithm. :param graph: The adjacency list of the graph where graph[u] contains (v, w) pairs. :param start: The starting node. :param end: The target node. :param n: The number of nodes. :return: The shortest path cost or 'UNREACHABLE'. distances = {node: float('infinity') for node in range(1, n+1)} distances[start] = 0 priority_queue = [(0, start)] while priority_queue: current_distance, current_node = heapq.heappop(priority_queue) if current_distance > distances[current_node]: continue for neighbor, weight in graph[current_node]: distance = current_distance + weight if distance < distances[neighbor]: distances[neighbor] = distance heapq.heappush(priority_queue, (distance, neighbor)) return distances[end] if distances[end] != float('infinity') else 'UNREACHABLE' def intra_galactic_transport(inputs): index = 0 results = [] while index < len(inputs): line = inputs[index] if line == '-1': break N, M = map(int, line.split()) index += 1 graph = {i: [] for i in range(1, N+1)} for _ in range(M): U, V, W = map(int, inputs[index].split()) graph[U].append((V, W)) graph[V].append((U, W)) # Assuming the routes are bidirectional index += 1 S, T = map(int, inputs[index].split()) index += 1 result = dijkstra(graph, S, T, N) results.append(str(result)) return results

answer:def longest_common_prefix(strings): Returns the length of the longest common prefix in a list of strings. Parameters: strings (list of str): A list of strings to find the longest common prefix. Returns: int: Length of the longest common prefix. if not strings: return 0 # Find the shortest string in the list shortest_str = min(strings, key=len) for i in range(len(shortest_str)): char = shortest_str[i] for string in strings: if string[i] != char: return i return len(shortest_str)

answer:def minimalBridgeCost(n, bridges): Returns the minimum cost required to connect all islands. :param n: int, the number of islands :param bridges: list of tuples, where each tuple contains three integers representing a bridge (island1, island2, weight) :return: int, minimum cost required to connect all islands parent = list(range(n)) rank = [0] * n def find(p): if parent[p] != p: parent[p] = find(parent[p]) return parent[p] def union(p1, p2): root1 = find(p1) root2 = find(p2) if root1 != root2: if rank[root1] > rank[root2]: parent[root2] = root1 elif rank[root1] < rank[root2]: parent[root1] = root2 else: parent[root2] = root1 rank[root1] += 1 # Sort bridges by their weight bridges.sort(key=lambda x: x[2]) mst_cost = 0 mst_edges = 0 for u, v, weight in bridges: if find(u) != find(v): union(u, v) mst_cost += weight mst_edges += 1 if mst_edges == n - 1: break return mst_cost