<p>You are given a positive integer <code>n</code> representing the number of nodes of a <strong>Directed Acyclic Graph</strong> (DAG). The nodes are numbered from <code>0</code> to <code>n - 1</code> (<strong>inclusive</strong>).</p>

<p>You are also given a 2D integer array <code>edges</code>, where <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>]</code> denotes that there is a <strong>unidirectional</strong> edge from <code>from<sub>i</sub></code> to <code>to<sub>i</sub></code> in the graph.</p>

<p>Return <em>a list</em> <code>answer</code><em>, where </em><code>answer[i]</code><em> is the <strong>list of ancestors</strong> of the</em> <code>i<sup>th</sup></code> <em>node, sorted in <strong>ascending order</strong></em>.</p>

<p>A node <code>u</code> is an <strong>ancestor</strong> of another node <code>v</code> if <code>u</code> can reach <code>v</code> via a set of edges.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/12/12/e1.png" style="width: 322px; height: 265px;" />
<pre>
<strong>Input:</strong> n = 8, edgeList = [[0,3],[0,4],[1,3],[2,4],[2,7],[3,5],[3,6],[3,7],[4,6]]
<strong>Output:</strong> [[],[],[],[0,1],[0,2],[0,1,3],[0,1,2,3,4],[0,1,2,3]]
<strong>Explanation:</strong>
The above diagram represents the input graph.
- Nodes 0, 1, and 2 do not have any ancestors.
- Node 3 has two ancestors 0 and 1.
- Node 4 has two ancestors 0 and 2.
- Node 5 has three ancestors 0, 1, and 3.
- Node 6 has five ancestors 0, 1, 2, 3, and 4.
- Node 7 has four ancestors 0, 1, 2, and 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/12/12/e2.png" style="width: 343px; height: 299px;" />
<pre>
<strong>Input:</strong> n = 5, edgeList = [[0,1],[0,2],[0,3],[0,4],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
<strong>Output:</strong> [[],[0],[0,1],[0,1,2],[0,1,2,3]]
<strong>Explanation:</strong>
The above diagram represents the input graph.
- Node 0 does not have any ancestor.
- Node 1 has one ancestor 0.
- Node 2 has two ancestors 0 and 1.
- Node 3 has three ancestors 0, 1, and 2.
- Node 4 has four ancestors 0, 1, 2, and 3.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= edges.length &lt;= min(2000, n * (n - 1) / 2)</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>
	<li>There are no duplicate edges.</li>
	<li>The graph is <strong>directed</strong> and <strong>acyclic</strong>.</li>
</ul>

class Solution {
public:
    vector<vector<int>> getAncestors(int n, vector<vector<int>>& edges) {
        vector<vector<int>> ans(n, vector<int>());

        helper(n, edges, ans);

        return ans;
    }

    void helper(int n, vector<vector<int>>& edges, vector<vector<int>>& ans)
    {
        vector<vector<int>> graph(n, vector<int>());
        unordered_map<int, vector<int>> ansSet;
        vector<bool> calculated(n, false);
       // vector<vector<int>> ansSet;

        for (auto edge : edges)
        {
            int from = edge[0];
            int to = edge[1];

            graph[to].push_back(from);
        }

        for (int i = 0; i < n; i++)
        {
            vector<bool> seen(n, false);
           // vector<int> vec;
           set<int> vec;
            dfs(i, graph, calculated,ans, vec, seen);
           
            if (vec.empty())
            {
                continue;
            }
           
          //  sort(vec.begin(), vec.end());
          //  auto it = unique(vec.begin(), vec.end());
          //  vec.erase(it, vec.end());
          vector<int> vec1(vec.begin(), vec.end());
            ans[i] = vec1;
            calculated[i] = 1;
           // ansSet[i] = vec;
        }

    }

    void dfs(int ind, vector<vector<int>>& graph, vector<bool>& calculated, vector<vector<int>>& ans, 
    set<int>& vec, vector<bool>& seen)
    {
        auto nodes = graph[ind];

        seen[ind] = 1;

        for (auto node : nodes)
        {
            if (seen[node])
            {
                continue;
            }

           // vec.push_back(node);
           vec.insert(node);
           
            if (calculated[node])
            {
                auto tempSet = ans[node];
                for (auto t : tempSet)
                {
                    if (seen[t])
                    {
                        continue;
                    }
                    seen[t]= 1;
                   // vec.push_back(t);
                   vec.insert(t);
                }
            }

            dfs(node, graph, calculated, ans, vec, seen);
        }
    }
};
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