There are a total of
numCourses courses you have to take, labeled from
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair:
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
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.
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.
- 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
class Solution: def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: graph = [ for i in range(numCourses)] in_degree =  * numCourses for to_node, from_node in prerequisites: graph[from_node].append(to_node) in_degree[to_node] += 1 num_choose = 0 queue = deque() for i in range(numCourses): if in_degree[i] == 0: queue.append(i) while queue: current_node = queue.popleft() num_choose += 1 for next_node in graph[current_node]: in_degree[next_node] -= 1 if in_degree[next_node] == 0: queue.append(next_node) return num_choose == numCourses
- Time Complexity: ~V + E
- Space Complexity: ~V + E
where V is the number of courses, and E is the number of dependencies.