<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 {
public:
    bool dfs(vector<vector<int>> &graph, vector<bool> &vis, unordered_set<int> &out, int node, vector<int> &ans)
    {
        //if visited? cycle detected. return false
        if(vis[node])
            return false;
        if(out.find(node) != out.end()) //already process can be skipped with true(since no cycle detected till now)
            return true;

        vis[node] = true;
        for(auto pre_req: graph[node])  //for every pre_requisite fo current node, recurse
        {
            if (dfs(graph, vis, out, pre_req, ans) == false)
                return false;
        }
        //node now processed, not can be added in out
        out.insert(node);
        ans.push_back(node);
        vis[node] = false; //backtrack
        return true; //since, no cycle detected till now
    }

    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        
        //buid adjacency list from pre-req
        vector<vector<int>> graph(numCourses);
        for (auto item: prerequisites)
        {
            graph[item[0]].push_back(item[1]);
        }
        
        //run dfs for every node
        vector<int> ans;  
        unordered_set<int> out;//this means covered in output, hence no need to process that node.
        vector<bool> vis(numCourses);  //node is visited and in the current recursion, to detect cycle.

        for (int node = 0; node < numCourses; node++)
            if(dfs(graph, vis, out, node, ans) == false)
                return {};

        return ans;
    }
};
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