210. Course Schedule II (Python)

Related Topic

Breadth-First-Search. Topological-Sort. Graph.

Description

There are a total of n courses you have to take, labeled from 0 to n-1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

Sample I/O

Example 1

Input: 2, [[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: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [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] .

Constrains

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

Methodology

This is a graph question. Any graph question to get directed acyclic graph (DAG) should think about using the topologic sort. This question is similar with question 207. Course Schedule. The only difference is now we are not just detecting if there is cycle. Moreover, we want to find one correct order of the courses if there are not cycles.

BFS:

def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        courses = collections.defaultdict(set)
        pres = collections.defaultdict(set)
        res = []
        for course, pre in prerequisites:
            courses[course].add(pre)
            pres[pre].add(course)
        stack=[n for n in range(numCourses) if not courses[n]]
        count = 0
        while stack:
            no_pre = stack.pop()
            res.append(no_pre)
            count+=1
            for course in pres[no_pre]:
                courses[course].remove(no_pre)
                if not courses[course]:
                    stack.append(course)
        return res if count==numCourses else []

BigO

Time Complexity: O(E+V) where V is the number of courses, and E is the number of dependencies.