--- title: '210. Course Schedule II' description: 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 sidebar: label: 'Course Schedule II' badge: 'Medium' --- Depth-First Search ### Example 1: - 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]`. ### Example 2: - 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]`. ### Example 3: - Input: `numCourses = 1, prerequisites = []` - Output: `[0]` ### Constraints: - `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. ## Solution ```py class Solution: def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]: prereq = {c: [] for c in range(numCourses)} for crs, pre in prerequisites: prereq[crs].append(pre) # course has 3 possible states: # visited -> crs has been added to output # visiting -> crs not added to output, but added to cycle # unvisited -> crs not added to output nor cycle output = [] visit, cycle = set(), set() def dfs(crs): if crs in cycle: return False if crs in visit: return True cycle.add(crs) for pre in prereq[crs]: if dfs(pre) == False: return False cycle.remove(crs) visit.add(crs) output.append(crs) return True for c in range(numCourses): if dfs(c) == False: return [] return output ```