<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 {
public:

     void dfs(int it, vector<int> &visited,
     vector<int> adj[],vector<int> &distance)
     {
             visited[it]=1;
             if(adj[it].size()==1 && visited[adj[it][0]]==0)
           {   distance[adj[it][0]] = distance[it] + 1;
            dfs(adj[it][0],visited,adj,distance);
           }
             return ;
     }
    int closestMeetingNode(vector<int>& edges, int node1, int node2) {
        int n=edges.size();
        int minDist = INT_MAX;
        vector<int> adj[n+1];
        for(int i=0;i<n;i++){
             if(edges[i]!=-1)
             {
                adj[i].push_back(edges[i]);
             }
        }
        
         vector<int> dist1(n, 0), dist2(n, 0);
        vector<int> visited1(n+3,0),visited2(n+3,0);
        if(adj[node1].size()==1 && adj[node2].size()==1)
        {   visited1[node1]=1;
         visited2[node2]=1;
            dfs(node1,visited1,adj,dist1);
            dfs(node2,visited2,adj,dist2);
              int ans=-1;
            for(int i=0;i<n;i++){
              
                if(visited1[i]==1 && visited2[i]==1 && 
                minDist > max(dist1[i], dist2[i])){
                
                
                    minDist = max(dist1[i], dist2[i]);
ans=i;
                  
                }
                 
            }
            return ans;
            
        }
        else if(adj[node1].size()==1)
        {   visited1[node1]=1;
            dfs(node1,visited1,adj,dist1);
            if(visited1[node2]==1)
            {
                return node2;
            }
            
        }
        else if(adj[node2].size()==1)
        { visited2[node2]=1;
            dfs(node2,visited2,adj,dist2);
             if(visited2[node1]==1)
            {
                return node1;
            }
        }
        
        if(node1==node2)
        {
              return node1;
        }
        
            return -1;
        


    }
};
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