Course Schedule II

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

• For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

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.

Example 1:
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].

Example 2:
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].

Example 3:
Input: numCourses = 1, prerequisites = []
Output: [0]

Constraints:

• 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.

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's prereq to course itself, then find the topological sorting of the graph.

Implementation

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 

Solution Implementation