<p>Given a directed acyclic graph (<strong>DAG</strong>) of <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>, find all possible paths from node <code>0</code> to node <code>n - 1</code> and return them in <strong>any order</strong>.</p>

<p>The graph is given as follows: <code>graph[i]</code> is a list of all nodes you can visit from node <code>i</code> (i.e., there is a directed edge from node <code>i</code> to node <code>graph[i][j]</code>).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/28/all_1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> graph = [[1,2],[3],[3],[]]
<strong>Output:</strong> [[0,1,3],[0,2,3]]
<strong>Explanation:</strong> There are two paths: 0 -&gt; 1 -&gt; 3 and 0 -&gt; 2 -&gt; 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/28/all_2.jpg" style="width: 423px; height: 301px;" />
<pre>
<strong>Input:</strong> graph = [[4,3,1],[3,2,4],[3],[4],[]]
<strong>Output:</strong> [[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]
</pre>

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

<ul>
	<li><code>n == graph.length</code></li>
	<li><code>2 &lt;= n &lt;= 15</code></li>
	<li><code>0 &lt;= graph[i][j] &lt; n</code></li>
	<li><code>graph[i][j] != i</code> (i.e., there will be no self-loops).</li>
	<li>All the elements of <code>graph[i]</code> are <strong>unique</strong>.</li>
	<li>The input graph is <strong>guaranteed</strong> to be a <strong>DAG</strong>.</li>
</ul>

class Solution {
public:
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        // using dfs
//         stack<vector<int>> s;
//         int n = graph.size();
//         vector<vector<int>> output;
//         s.push({0});
        
//         while(!s.empty())
//         {
//             vector<int> curr_path = s.top();
//             s.pop();
//             int curr_size = curr_path.size();
//             if(curr_path[curr_size - 1] == n-1)
//             {
//                 output.push_back(curr_path);
//                 continue;
//             }
//             curr_path.push_back(-1);
//             for(auto neigh:graph[curr_path[curr_size - 1]])
//             {
//                 curr_path[curr_path.size() - 1] = neigh;
//                 s.push(curr_path);
//             }
//         }
        
//         return output;
        
        // using bfs
        queue<vector<int>> q;
        int n = graph.size();
        vector<vector<int>> output;
        q.push({0});
        
        while(!q.empty())
        {
            vector<int> curr_path = q.front();
            q.pop();
            if(curr_path[curr_path.size() - 1] == n-1)
            {
                output.push_back(curr_path);
                continue;
            }
            
            for(auto neigh:graph[curr_path[curr_path.size() - 1]])
            {
                curr_path.push_back(neigh);
                q.push(curr_path);
                curr_path.pop_back();
            }
        }
        
        return output;
    }
};
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