You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given an array relations where relations[i] = [prevCourse_i, nextCourse_i], representing a prerequisite relationship between course prevCourse_i and course nextCourse_i: course prevCourse_i has to be taken before course nextCourse_i.In one semester, you can take any number of courses as long as you have taken all the prerequisites in the previous semester for the courses you are taking.Return the minimum number of semesters needed to take all courses. If there is no way to take all the courses, return -1.
from collections import dequeclass Solution: # Time: O(n + m) # Space: O(n + m) def minimum_semesters(self, n: int, relations: list[list[int]]) -> int: graph: list[list[int]] = [[] for _ in range(n)] indegree = [0] * n for prev_course, next_course in relations: graph[prev_course - 1].append(next_course - 1) indegree[next_course - 1] += 1 queue: deque[int] = deque(i for i in range(n) if indegree[i] == 0) semesters = 0 taken = 0 while queue: semesters += 1 for _ in range(len(queue)): course = queue.popleft() taken += 1 for nxt in graph[course]: indegree[nxt] -= 1 if indegree[nxt] == 0: queue.append(nxt) return semesters if taken == n else -1
