There are a total of
numCourses courses you have to take, labeled from
numCourses - 1. You are given an array
prerequisites[i] = [ai, bi] indicates that you must take course
bi first if you want to take course
- For example, the pair
[0, 1], indicates that to take course
0you have to first take course
Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.
Input: numCourses = 2, prerequisites = [[1,0]] Output: [0,1] Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].
Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]] Output: [0,2,1,3] Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].
Input: numCourses = 1, prerequisites =  Output: 
1 <= numCourses <= 2000
0 <= prerequisites.length <= numCourses * (numCourses - 1)
prerequisites[i].length == 2
0 <= ai, bi < numCourses
ai != bi
- All the pairs
[ai, bi]are distinct.
Similar to the problem Course Schedule I, we can use topological sort to solve this problem. Base on the indegree values for the courses to find out the result.
class Solution: def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]: results =  adjacent_list =  indegrees =  for i in range(numCourses): adjacent_list.append() indegrees.append(0) for prerequisite in prerequisites: adjacent_list[prerequisite].append(prerequisite) indegrees[prerequisite] += 1 queue =  for i, indegree in enumerate(indegrees): if indegree == 0: queue.append(i) while queue: course = queue.pop(0) results.append(course) for adj_course in adjacent_list[course]: indegrees[adj_course] -= 1 if indegrees[adj_course] == 0: queue.append(adj_course) if len(results) < numCourses: return  return results
- Time Complexity: ~V + E
- Space Complexity: ~V + E
where V is the number of courses, and E is the number of prerequisites.