leetcode/apps/docs/content/(04-relational)/topological-sort/210-course-schedule-ii.mdx

---
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
icon: line-dot-left-horizontal
sidebar:
  label: 'Course Schedule II'
  badge: 'Medium'
---

<Badge variant="accent">Depth-First Search</Badge>

### 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
```