<p>In this problem, a tree is an <strong>undirected graph</strong> that is connected and has no cycles.</p>

<p>You are given a graph that started as a tree with <code>n</code> nodes labeled from <code>1</code> to <code>n</code>, with one additional edge added. The added edge has two <strong>different</strong> vertices chosen from <code>1</code> to <code>n</code>, and was not an edge that already existed. The graph is represented as an array <code>edges</code> of length <code>n</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the graph.</p>

<p>Return <em>an edge that can be removed so that the resulting graph is a tree of </em><code>n</code><em> nodes</em>. If there are multiple answers, return the answer that occurs last in the input.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/02/reduntant1-1-graph.jpg" style="width: 222px; height: 222px;" />
<pre>
<strong>Input:</strong> edges = [[1,2],[1,3],[2,3]]
<strong>Output:</strong> [2,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/02/reduntant1-2-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
<strong>Input:</strong> edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
<strong>Output:</strong> [1,4]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == edges.length</code></li>
	<li><code>3 &lt;= n &lt;= 1000</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt;= edges.length</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated edges.</li>
	<li>The given graph is connected.</li>
</ul>

class Solution {
public:
    stack<int> maxes;
    unordered_set<int> visiteds;
    unordered_set<int> notGoingToVisits;
    int dfs(int idx,int fromIdx){
        auto it2=notGoingToVisits.find(idx);
        if(it2!=notGoingToVisits.end()) return -1;
        auto it =visiteds.find(idx);
        if(it!=visiteds.end())return idx;
        visiteds.emplace_hint(it,idx);
        for(auto&[neighbor,w]:es[idx]){
            if(fromIdx!=neighbor){
                if(maxes.empty()) maxes.emplace(w);
                else maxes.emplace(max(maxes.top(),w));
                int result=dfs(neighbor,idx);
                if(result!=-1) return result;
                else {
                    maxes.pop();
                    notGoingToVisits.emplace(neighbor);
                }
            }
        }
        return -1;
    }
    unordered_map<int,unordered_map<int,int>> es;
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        for(int i=0;i<edges.size();i++){
            es[edges[i][0]].emplace(edges[i][1],i);
            es[edges[i][1]].emplace(edges[i][0],i);
        }
        
        int node= dfs(1,-1);
        visiteds.clear();
        while(!maxes.empty())maxes.pop();
        dfs(node,-1);
        return edges[maxes.top()];
    }
};
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