# course-schedule-ii


Try it on <a href='https://leetcode.com/problems/course-schedule-ii'>leetcode</a>

## Description
<div class="description">
<div><p>There are a total of <code>numCourses</code> courses you have to take, labeled from <code>0</code> to <code>numCourses - 1</code>. You are given an array <code>prerequisites</code> where <code>prerequisites[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that you <strong>must</strong> take course <code>b<sub>i</sub></code> first if you want to take course <code>a<sub>i</sub></code>.</p>

<ul>
	<li>For example, the pair <code>[0, 1]</code>, indicates that to take course <code>0</code> you have to first take course <code>1</code>.</li>
</ul>

<p>Return <em>the ordering of courses you should take to finish all courses</em>. If there are many valid answers, return <strong>any</strong> of them. If it is impossible to finish all courses, return <strong>an empty array</strong>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input:</strong> numCourses = 2, prerequisites = [[1,0]]
<strong>Output:</strong> [0,1]
<strong>Explanation:</strong> 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].
</pre>

<p><strong>Example 2:</strong></p>

<pre><strong>Input:</strong> numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
<strong>Output:</strong> [0,2,1,3]
<strong>Explanation:</strong> 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].
</pre>

<p><strong>Example 3:</strong></p>

<pre><strong>Input:</strong> numCourses = 1, prerequisites = []
<strong>Output:</strong> [0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= numCourses &lt;= 2000</code></li>
	<li><code>0 &lt;= prerequisites.length &lt;= numCourses * (numCourses - 1)</code></li>
	<li><code>prerequisites[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; numCourses</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>All the pairs <code>[a<sub>i</sub>, b<sub>i</sub>]</code> are <strong>distinct</strong>.</li>
</ul>
</div>
</div>

## Solution(Python)
```Python
from collections import defaultdict, deque


class Solution:
    def findOrder(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: List[int]
        """

        # Prepare the graph
        adj_list = defaultdict(list)
        indegree = {}
        for dest, src in prerequisites:
            adj_list[src].append(dest)

            # Record each node's in-degree
            indegree[dest] = indegree.get(dest, 0) + 1

        # Queue for maintainig list of nodes that have 0 in-degree
        zero_indegree_queue = deque(
            [k for k in range(numCourses) if k not in indegree])

        topological_sorted_order = []

        # Until there are nodes in the Q
        while zero_indegree_queue:

            # Pop one node with 0 in-degree
            vertex = zero_indegree_queue.popleft()
            topological_sorted_order.append(vertex)

            # Reduce in-degree for all the neighbors
            if vertex in adj_list:
                for neighbor in adj_list[vertex]:
                    indegree[neighbor] -= 1

                    # Add neighbor to Q if in-degree becomes 0
                    if indegree[neighbor] == 0:
                        zero_indegree_queue.append(neighbor)

        return (
            topological_sorted_order
            if len(topological_sorted_order) == numCourses
            else []
        )

```