<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 {
    // Function for Depth First Search
    void dfs(int i, vector<bool>& visited, vector<vector<int>>& adj, vector<int>& distance_node1, int dist)
    {
        // Marking the node visited
        visited[i]=true;
        // Storing the distance 
        distance_node1[i]=dist;
        // Checking the further nodes
        for (auto it: adj[i]) if (!visited[it]) dfs(it, visited, adj, distance_node1, dist+1);
    }
public:
    int closestMeetingNode(vector<int>& edges, int node1, int node2) {
        // Data structure to store Distance of nodes from node1 & node 2
        vector<int> distance_node1(edges.size(),1e6);
        vector<int> distance_node2(edges.size(),1e6);
        // Visited array to keep a traack of visited nodes
        vector<bool> visited(edges.size(),false);
        // Declaring Adjacency List
        vector<vector<int>> adj(edges.size());

        // Filling Adjacency List
        for (int i=0;i<edges.size();i++)
        {
            // Treating -1 as self loop
            if (edges[i]==-1) adj[i].push_back(i);
            else adj[i].push_back(edges[i]);
        }
        // Calling dfs to fill the distance needed for each node
        dfs(node1, visited, adj, distance_node1, 0);
        // Reset the visited array after first dfs call
        for (int i=0;i<edges.size();i++) visited[i]=false;
        // Calling dfs to fill the distance needed for each node
        dfs(node2, visited, adj, distance_node2, 0);

        int ans=1e6;
        // Keeping track of index
        int ind=-1;
        for (int i=0;i<edges.size();i++)
        {
            // Minimising the maximum distance of node from node1 & node2
            if (ans>max(distance_node1[i],distance_node2[i]))
            {
                ans=max(distance_node1[i],distance_node2[i]);
                ind=i;
            }
        }
        // Return the index
        return ind;
    }
};
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