<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 {
    void dfs(vector<vector<int>>& graph, vector<int> &mp, vector<bool>& vis, int node, int w){
        mp[node] = w;
        vis[node] = true;
        for(int i = 0; i < graph[node].size(); i++){
            int child = graph[node][i];
            if(vis[child] == false || w + 1 < mp[child]) dfs(graph, mp, vis, child, w + 1);
        }
    }
public:
    int closestMeetingNode(vector<int>& edges, int node1, int node2) {
        int n = edges.size();
        vector<vector<int>> graph(n);
        for(int i = 0; i < n; i++){
            if(edges[i] == -1) continue;
            int sec = edges[i];
            graph[i].push_back(sec);
        }
        vector<int> mp1(n, -1);
        vector<int> mp2(n, -1);
        vector<bool> vis1(n, false);
        dfs(graph, mp1, vis1, node1, 0);
        vector<bool> vis2(n, false);
        dfs(graph, mp2, vis2, node2, 0);
        int mx = INT_MAX;
        int node = -1;
        for(int i = 0; i < n; i++){
            if(mp1[i] != -1 && mp2[i] != -1) {
                int t = max(mp1[i], mp2[i]);
                if(t < mx) {
                    mx = t;
                    node = i;
                }
            }
        }
        return node;
    }
};

/*
class Solution {
    void dfs(vector<int>& edges, vector<int> &mp, vector<bool>& vis, int node, int w){
        vis[node] = true;  
        mp[node] = w;
        int child = edges[node];
        if (child != -1 && (!vis[child] || w + 1 < mp[child])) {
            dfs(edges, mp, vis, child, w + 1);
        }
    }

public:
    int closestMeetingNode(vector<int>& edges, int node1, int node2) {
        int n = edges.size();

        vector<int> mp1(n, -1);  
        vector<int> mp2(n, -1);  
        vector<bool> vis1(n, false);  
        vector<bool> vis2(n, false);  

        
        dfs(edges, mp1, vis1, node1, 0);
        dfs(edges, mp2, vis2, node2, 0);

        int minDistance = INT_MAX;
        int resultNode = -1;

        for (int i = 0; i < n; i++) {
            if (mp1[i] != -1 && mp2[i] != -1) {
                int maxDist = max(mp1[i], mp2[i]);
                if (maxDist < minDistance) {
                    minDistance = maxDist;
                    resultNode = i;
                }
            }
        }

        return resultNode;
    }
};
*/
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