207. Course Schedule
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, is it possible for you to finish all courses?
Example:
2, [[1,0]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.
2, [[1,0],[0,1]]
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.
Note:
- 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.
Hints:
- 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.
- Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
- Topological sort could also be done via BFS.
Solution: Topological Sort - DFS
class Solution(object):
def canFinish(self, numCourses, prerequisites):
"""
:type numCourses: int
:type prerequisites: List[List[int]]
:rtype: bool
"""
graph = [[] for _ in xrange(numCourses)]
for cur, pre in prerequisites:
graph[pre].append(cur)
return not self.has_cycle(graph)
def has_cycle(self, graph):
visited = set()
visiting = set()
cycle = []
def dfs(node):
if cycle:
return True
if node in visiting:
cycle.append(1)
return True
if node in visited:
return False
visiting.add(node)
for neighbor in graph[node]:
if dfs(neighbor):
return True
visiting.remove(node)
visited.add(node)
return False
return any(dfs(i) for i in xrange(len(graph)))