<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:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        vector<int> ans; 
        set<int> s; 
        for(auto &x:edges)
        {
            s.insert(x[0]); s.insert(x[1]); 
        }
        int cnt; 
        int n = s.size(); 
        for(int i = edges.size()-1; i >= 0; i-- )
        {
             
            unordered_map<int, vector<int>> mp; 
            for(int j = edges.size()-1;j>=0;j--)
            {
                if(i!=j)
                {
                    mp[edges[j][0]-1].push_back(edges[j][1]-1); 
                    mp[edges[j][1]-1].push_back(edges[j][0]-1); 
                }
            }
            vector<int> visited(n,0);
            cnt = 0; 

            queue<int> qu; 
            qu.push(0); 
            visited[0] = 1; 
            cnt++; 
            while(!qu.empty())
            {
                int node = qu.front(); 
                qu.pop(); 

                for(auto &x : mp[node])
                {
                    if(!visited[x])
                    {
                        visited[x] = 1; 
                        qu.push(x); 
                        cnt++; 
                    }
                }
            } 

            if(cnt == n)
            {
                ans.push_back(edges[i][0]);
                ans.push_back(edges[i][1]);
                break; 
            }
        }    
        return ans; 
    }
};
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