<p>You are given a <strong>directed</strong> graph of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>, where each node has <strong>at most one</strong> outgoing edge.</p>

<p>The graph is represented with a given <strong>0-indexed</strong> array <code>edges</code> of size <code>n</code>, indicating that there is a directed edge from node <code>i</code> to node <code>edges[i]</code>. If there is no outgoing edge from <code>i</code>, then <code>edges[i] == -1</code>.</p>

<p>You are also given two integers <code>node1</code> and <code>node2</code>.</p>

<p>Return <em>the <strong>index</strong> of the node that can be reached from both </em><code>node1</code><em> and </em><code>node2</code><em>, such that the <strong>maximum</strong> between the distance from </em><code>node1</code><em> to that node, and from </em><code>node2</code><em> to that node is <strong>minimized</strong></em>. If there are multiple answers, return the node with the <strong>smallest</strong> index, and if no possible answer exists, return <code>-1</code>.</p>

<p>Note that <code>edges</code> may contain cycles.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-2.png" style="width: 321px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [2,2,3,-1], node1 = 0, node2 = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> The distance from node 0 to node 2 is 1, and the distance from node 1 to node 2 is 1.
The maximum of those two distances is 1. It can be proven that we cannot get a node with a smaller maximum distance than 1, so we return node 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-4.png" style="width: 195px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [1,2,-1], node1 = 0, node2 = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> The distance from node 0 to node 2 is 2, and the distance from node 2 to itself is 0.
The maximum of those two distances is 2. It can be proven that we cannot get a node with a smaller maximum distance than 2, so we return node 2.
</pre>

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

<ul>
	<li><code>n == edges.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-1 &lt;= edges[i] &lt; n</code></li>
	<li><code>edges[i] != i</code></li>
	<li><code>0 &lt;= node1, node2 &lt; n</code></li>
</ul>

class Solution {
private: 
    bool check; 
    int node; 
public:

    void dfs( int root, vector<bool>& visited, vector<int>& edges, vector<int>& result )
    {
        if( root == node )
            check = true; 
        visited[root] = true; 
        result.push_back( root );

        if( edges[root] != -1 && !visited[edges[root]] )
        dfs( edges[root], visited, edges, result ); 
    }

    int closestMeetingNode(vector<int>& edges, int node1, int node2) {
        int n = edges.size(); 
        vector<bool> visited(n , false); 

        check = false; 
        vector<int> result; 
        node = node2; 
        dfs( node1 , visited, edges, result ); 
        
        unordered_map<int, int> mp; 
        for( int i = 0; i < result.size(); i++ )
        {
            mp[result[i]] = i; 
        }

        for( int i = 0; i < n; i++ )
            visited[i] = false; 

        vector<int> new_res; 
        node = node1; 
        dfs( node2, visited, edges, new_res); 


        // for( auto ele: result )
        //     cout << ele << " "; 
        // cout << endl; 

        // for( auto ele: new_res )
        //     cout << ele << " "; 
        // cout << endl; 


        vector<int> dist(n, INT_MAX ); 
        for( int i = 0; i < new_res.size(); i++ )
        {
            if( mp.find( new_res[i] ) == mp.end() )
                continue;
            dist[new_res[i]] = min( dist[new_res[i]], max( i, mp[new_res[i]]) ); 
        }

        int idx = 0;
        int ans = INT_MAX;
        for( int i = 0; i < n; i++ )
        {
            if( ans > dist[i] )
            {
                ans = dist[i];  
                idx = i; 
            }
        }

        if( ans == INT_MAX )
            return -1; 
        return idx; 
    }
};
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