Description
https://leetcode.com/problems/course-schedule/
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 course0
you have to first take course1
.
Return true
if you can finish all courses. Otherwise, return false
.
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:
1 <= numCourses <= 105
0 <= prerequisites.length <= 5000
prerequisites[i].length == 2
0 <= ai, bi < numCourses
- All the pairs prerequisites[i] are unique.
Explanation
Use topological sort to keep appending the course with indegree 0 to the queue. In the end, check how many number of courses can be taken.
Python Solution
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
adjacency_list = []
indegrees = []
for i in range(numCourses):
adjacency_list.append([])
indegrees.append(0)
for prerequisite in prerequisites:
adjacency_list[prerequisite[0]].append(prerequisite[1])
indegrees[prerequisite[1]] += 1
queue = []
for i, indegree in enumerate(indegrees):
if indegree == 0:
queue.append(i)
count = 0
while queue:
course = queue.pop(0)
count += 1
for adj_course in adjacency_list[course]:
indegrees[adj_course] -= 1
if indegrees[adj_course] == 0:
queue.append(adj_course)
return count == numCourses
- Time Complexity: ~V + E
- Space Complexity: ~V + E
where V is the number of courses, and E is the number of dependencies.