LeetCode Course Schedule II Solution
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.
numCourses = 2, prerequisites = [[1,0]]
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].
numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
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].
numCourses = 1, prerequisites = 
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.
Problem Link: https://leetcode.com/problems/course-schedule-ii/
The problems asks us to find the ordering of courses to take. We know this involves the dependency of courses, so we will find the topological ordering of these courses.
We will build a graph with directed edges from a
course itself, then find the topological sorting of the graph.
1def findOrder(numCourses: int, prerequisites: List[List[int]]) -> List[int]: 2 visited = [False for _ in range(numCourses)] 3 finished =  4 graph = defaultdict(list) 5 6 for course, prereq in prerequisites: 7 graph[prereq].append(course) 8 9 def dfs(prereq): 10 for course in graph[prereq]: 11 if not visited[course]: 12 visited[course] = True 13 if not dfs(course): return False # unable to finish course 14 elif course not in finished: # cycle in graph 15 return False 16 finished.append(prereq) 17 return True 18 19 for i in range(numCourses): # visit all trees in graph 20 if not visited[i]: 21 visited[i] = True 22 if not dfs(i): return  23 24 finished.reverse() # topological ordering = reverse finish time 25 return finished