<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 class="example">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 class="example">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 class="example">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>

class Solution:
    def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
        adj = {i:[] for i in range(numCourses)}
        for x, y in prerequisites:
            adj[y].append(x)
        visited, cur = set(), set()
        order = []
        def dfs(x: int):
            if x in visited:
                return True
            cur.add(x)
            visited.add(x)
            for dx in adj[x]:
                if dx in cur or not dfs(dx):
                    return False
            cur.discard(x)
            order.append(x)
            return True
        
        for x in range(numCourses):
            if not dfs(x):
                return []
        order.reverse()
        return order
        























        # n = numCourses
        # adj = {i:[] for i in range(n)}
        # for a,b in prerequisites:
        #     adj[b].append(a)
        # def dfs(node):
        #     if node in visited:
        #         return True
        #     visited.add(node)
        #     cur.add(node)
        #     for v in adj[node]:
        #         if v in cur or not dfs(v):
        #             return False
        #     cur.discard(node)
        #     res.append(node)
        #     return True
        # res = []
        # visited, cur = set(), set()
        # for i in range(n):
        #     if not dfs(i):
        #         return []
        # res.reverse()
        # return res











        # n = numCourses
        # adj = {i:[] for i in range(n)}
        # for u,v in prerequisites:
        #     adj[v].append(u)
        # visited, cur = set(), set()
        # order = []
        # def dfs(node):
        #     if node in visited:
        #         return True
        #     visited.add(node)
        #     cur.add(node)
        #     for u in adj[node]:
        #         if u in cur or not dfs(u):
        #             return False                
        #     cur.discard(node)
        #     order.append(node)
        #     return True
        # for i in range(n):
        #     if not dfs(i):
        #         return []
        # order.reverse()
        # return order

        
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