<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 {
private:
    /*void dfs (int source, int target, vector<int>& path, vector<vector<int>>& res, vector<vector<int>>& graph, vector<bool>& visited){
        visited[source] = true;
        path.push_back(source);
        if(source == target){
            res.push_back(path);
            path.pop_back();
            visited[source] = false;
            return;
        }

        for(int i = 0; i < graph[source].size(); ++i){
            if(!visited[graph[source][i]]){
                dfs(graph[source][i], target, path, res, graph, visited);
            }
        }
        visited[source] = false;
        path.pop_back();
    }*/

    vector<vector<int>> dfs (int source, int target, vector<vector<int>>& graph, vector<vector<vector<int>>>& memo){
        if(!memo[source].empty()){
            return memo[source];
        }

        if(source == target) return {{target}};
        vector<vector<int>> res;

        for(int i = 0; i < graph[source].size(); ++i){
            vector<vector<int>> tmp = dfs(graph[source][i], target, graph, memo);
            for(int i = 0; i < tmp.size(); ++i){
                tmp[i].insert(tmp[i].begin(), source);
                res.push_back(tmp[i]);
            }
        }
        memo[source] = res;
        return res;
    }

public:
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        /*int target = graph.size() - 1;
        vector<bool> visited(target + 1, false);
        vector<vector<int>> result;

        vector<int> path;
        dfs(0, target, path, result, graph, visited);
        return result;
        //O(2^n) + O(n)*/

        int target = graph.size() - 1;
        vector<vector<vector<int>>> memo (target + 1);
        return dfs(0, target, graph, memo);
    }
};
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