# 210. Course Schedule II

## Medium

***

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**.

```python
class Solution:
    def findOrder(self, N, P):
        G, indegree, q, ans = defaultdict(list), [0]*N, deque(), []
        for nxt, pre in P:
            G[pre].append(nxt)
            indegree[nxt] += 1
        
        for i in range(N):
            if indegree[i] == 0:
                q.append(i)
        while q:
            cur = q.popleft()
            ans.append(cur)
            for nextCourse in G[cur]:
                indegree[nextCourse] -= 1
                if indegree[nextCourse] == 0: 
                    q.append(nextCourse)
                    
        return ans if len(ans) == N else []
```