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97 lines
2.6 KiB
Python
97 lines
2.6 KiB
Python
"""
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210. Course Schedule II
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Difficulty: Medium
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https://leetcode.com/problems/course-schedule-ii/
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──────────────────────────────────────────────────
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There are a total of numCourses courses you have to take, labeled
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from 0 to numCourses - 1. You are given an array prerequisites where
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prerequisites[i] = [ai, bi] indicates that you must take course bi
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first if you want to take course ai.
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• For example, the pair [0, 1], indicates that to take course 0 you
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have to first take course 1.
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Return the ordering of courses you should take to finish all courses.
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If there are many valid answers, return any of them. If it is
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impossible to finish all courses, return an empty array.
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Example 1:
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Input: numCourses = 2, prerequisites = [[1,0]]
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Output: [0,1]
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Explanation: There are a total of 2 courses to take. To take course 1
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you should have finished course 0. So the correct course order is
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[0,1].
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Example 2:
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Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
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Output: [0,2,1,3]
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Explanation: There are a total of 4 courses to take. To take course 3
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you should have finished both courses 1 and 2. Both courses 1 and 2
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should be taken after you finished course 0.
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So one correct course order is [0,1,2,3]. Another correct ordering is
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[0,2,1,3].
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Example 3:
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Input: numCourses = 1, prerequisites = []
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Output: [0]
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Constraints:
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• 1 <= numCourses <= 2000
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• 0 <= prerequisites.length <= numCourses * (numCourses - 1)
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• prerequisites[i].length == 2
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• 0 <= ai, bi < numCourses
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• ai != bi
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• All the pairs [ai, bi] are distinct.
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"""
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class Solution:
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def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]:
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prereq = {c: [] for c in range(numCourses)}
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for crs, pre in prerequisites:
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prereq[crs].append(pre)
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# course has 3 possible states:
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# visited -> crs has been added to output
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# visiting -> crs not added to output, but added to cycle
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# unvisited -> crs not added to output nor cycle
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output = []
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visit, cycle = set(), set()
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def dfs(crs):
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if crs in cycle:
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return False
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if crs in visit:
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return True
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cycle.add(crs)
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for pre in prereq[crs]:
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if dfs(pre) == False:
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return False
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cycle.remove(crs)
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visit.add(crs)
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output.append(crs)
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return True
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for c in range(numCourses):
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if dfs(c) == False:
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return []
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return output
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