# 1136 Parallel Courses

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] = [prevCoursei, nextCoursei]`, representing a prerequisite relationship between course `prevCoursei` and course `nextCoursei`: course `prevCoursei` has to be taken before course `nextCoursei`.

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`.

Example 1:

``````1   2
\ /
v
3

Input: n = 3, relations = [[1,3],[2,3]]
Output: 2
Explanation: The figure above represents the given graph.
In the first semester, you can take courses 1 and 2.
In the second semester, you can take course 3.
``````

Example 2:

``````1 --> 2
^    /
\  v
3

Input: n = 3, relations = [[1,2],[2,3],[3,1]]
Output: -1
Explanation: No course can be studied because they are prerequisites of each other.
``````

Constraints:

• `1 <= n <= 5000`
• `1 <= relations.length <= 5000`
• `relations[i].length == 2`
• `1 <= prevCoursei, nextCoursei <= n`
• `prevCoursei != nextCoursei`
• All the pairs `[prevCoursei, nextCoursei]` are unique.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 `````` ``````class Solution: def minimumSemesters(self, n: int, relations: List[List[int]]) -> int: graph = defaultdict(list) indegree = defaultdict(int) for pre, course in relations: graph[pre].append(course) indegree[course] += 1 queue = [i for i in range(1, n+1) if i not in indegree] count = 0 while queue: count += 1 temp = [] for course in queue: for next_course in graph[course]: indegree[next_course] -= 1 if indegree[next_course] == 0: temp.append(next_course) del indegree[next_course] queue = temp return count if not indegree else -1``````