leetcode/work/1/Medium/Depth-First Search/210.course-schedule-ii.py

"""
210. Course Schedule II
Difficulty: Medium
https://leetcode.com/problems/course-schedule-ii/

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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 course 0 you
have to first take course 1.

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.



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.
"""


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