<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(vector<int>& edges, int curr, vector<int>& vis, int step, unordered_map<int, int> &mp)
    // {
    //     if (curr == -1) return;
        
    //     if (!vis[curr])
    //     {
    //         if (mp.find(curr) != mp.end()) mp[curr] = min(step, mp[curr]);
    //         else mp[curr] = step;
    //         vis[curr] = 1;
    //         dfs(edges, edges[curr], vis, step+1, mp);
    //     }
    // }
    void bfs(vector<int>& edges, int curr, vector<int>& vis, int step, unordered_map<int, int> &mp)
    {
        queue<int> q;
        q.push(curr);
        vis[curr] = 1;
        mp[curr] = step;
        step++;
        while (!q.empty())
        {
            int node = q.front();
            q.pop();
            int tmp = edges[node];
            if (tmp != -1 && !vis[tmp])
            {
                vis[tmp] = 1;
                if (mp.find(tmp) != mp.end()) mp[tmp] = min(step, mp[tmp]);
                else mp[tmp] = step;
                step++;
                q.push(tmp);
            }
        }
    }
    int closestMeetingNode(vector<int>& edges, int node1, int node2) {
        // int n = edges.size();
        // if (n == 3 && node1 == 2 && node2 == 0) return 0;
        vector<int> vis(edges.size(), 0);
        unordered_map<int, int> mp1, mp2;
        bfs(edges, node1, vis, 0, mp1);
        fill(vis.begin(), vis.end(), 0);
        bfs(edges, node2, vis, 0, mp2);
        int ans = INT_MAX, dis = INT_MAX;
        for (auto & a: mp1)
        {
            
            if (mp2.find(a.first) != mp2.end())
            {
                int tmpDis = max(mp2[a.first], a.second);
                cout<<a.first<<" "<<a.second<<"|| ";
                if (dis >= tmpDis)
                {
                    if (dis == tmpDis) ans = min(a.first, ans);
                    else ans = a.first;
                    dis = tmpDis;
                }
            }
        }
        return ans == INT_MAX ? -1 : ans;
    }
};
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