Day-29_Course_Schedule.py

'''
	There are a total of numCourses courses you have to take, labeled from 0 to numCourses-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, is it possible for you to finish all courses?

 

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.
Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0, and to take course 0 you should
             also have finished course 1. So it is impossible.
 

Constraints:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.
1 <= numCourses <= 10^5
   Hide Hint #1  
This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
   Hide Hint #2  
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
   Hide Hint #3  
Topological sort could also be done via BFS.




'''
class Solution:
  def canFinish(self, N: int, prereqs: List[List[int]]) -> bool:
    graph = defaultdict(list)
    for a,b in prereqs: graph[a].append(b)

    status = defaultdict(int)
    def dfs(i):
        if status[i]: return status[i] == 1
        status[i] = -1
        if not all(map(dfs, graph.get(i,()))): return False
        status[i] = 1
        return True

    return all(map(dfs, graph))