<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:
    vector <int> dijkstra(int V, vector<int> adj[], int S)
    {
        set<pair<int,int>> st; 
        vector<int> dist(V, 1e9); 
        st.insert({0, S}); 
        dist[S] = 0;
        while(!st.empty()) {
            auto it = *(st.begin()); 
            int node = it.second; 
            int dis = it.first; 
            st.erase(it); 
            for(auto it : adj[node]) {
                int adjNode = it;  
                if(dis + 1 < dist[adjNode]) {
                    if(dist[adjNode] != 1e9) 
                        st.erase({dist[adjNode], adjNode}); 
                    dist[adjNode] = dis + 1; 
                    st.insert({dist[adjNode], adjNode}); 
                 }
            }
        }
        return dist; 
    }
    int closestMeetingNode(vector<int>& edges, int node1, int node2) {
        int n=edges.size();
        vector<int>adj[n];
        for(int i=0;i<n;i++){
            if(edges[i]!=-1){
                adj[i].push_back(edges[i]);
            }
        }
        vector<int>distnode1,distnode2;
        distnode1=dijkstra(n,adj,node1);
        distnode2=dijkstra(n,adj,node2);
        for(auto i : distnode1){
           cout<<i<< " ";
        }        
        cout<<endl;
        for(auto i : distnode2){
           cout<<i<< " ";
        }        
       int ans=1e9;
        int ind=-1;
        for(int i=0;i<n;++i)
        {
            if(max(distnode1[i],distnode2[i])<ans)
            {
                ind=i;
            }
            ans=min(max(distnode1[i],distnode2[i]),ans);

        }
        return ind;
    }
};
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