There are a total of
n courses you have to take labelled from
n - 1.
Some courses may have
prerequisites, for example, if
prerequisites[i] = [ai, bi] this means you must take the course
bi before the course
Given the total number of courses
numCourses and a list of the
prerequisite pairs, return the ordering of courses you should take to finish all courses.
If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.
Input: numCourses = 2, prerequisites = [[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].
Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]] Output: [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].
Input: numCourses = 1, prerequisites =  Output: 
1 <= numCourses <= 2000
0 <= prerequisites.length <= numCourses * (numCourses - 1)
prerequisites[i].length == 2
0 <= ai, bi < numCourses
ai != bi
- All the pairs
[ai, bi]are distinct.
class Solution: def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]: graph = [ for i in range(numCourses)] in_degree =  * numCourses for node_in, node_out in prerequisites: graph[node_out].append(node_in) in_degree[node_in] += 1 num_choose = 0 queue = deque() topo_order =  for i in range(numCourses): if in_degree[i] == 0: queue.append(i) while queue: current = queue.popleft() topo_order.append(current) num_choose += 1 for next in graph[current]: in_degree[next] -= 1 if in_degree[next] == 0: queue.append(next) if num_choose == numCourses: return topo_order return 
- Time Complexity: ~V + E
- Space Complexity: ~V + E