<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 {

    unordered_map<int, unordered_set<int>> adj_map; 
    vector<bool> vis, all_vis, all_not_vis;

    bool has_cycle (int parent, int v)
    {
        vis[v] = true;
        for (auto w: adj_map[v])
        {
            if (!vis[w])
            {
                if (has_cycle (v, w))
                {
                    return true;
                }
            }
            else if (w != parent)
            {
                return true;
            }
        }

        return false;
    }

public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        
        int max_n = 0;
        for (int i = 0; i < edges.size(); i++)
        {
            adj_map[edges[i][0]].insert (edges[i][1]);
            adj_map[edges[i][1]].insert (edges[i][0]);
        }

        for (auto it : adj_map)
        {
            max_n = max (max_n, it.first);
        }
        all_not_vis = vector<bool>(max_n+1, false);
        all_not_vis[0] = true;
        vis = all_not_vis;
        
        all_vis = vector<bool>(max_n+1, true);

        for (int i = edges.size() - 1; i >= 0; i--)
        {
            adj_map[edges[i][0]].erase ({edges[i][1]});
            adj_map[edges[i][1]].erase ({edges[i][0]});
            //cout<<"Removed ("<<edges[i][0]<<","<<edges[i][1]<<")"<<endl<<"\t";

            if (!has_cycle (-1, 1))
            {
                if (vis == all_vis)
                    return {edges[i][0], edges[i][1]};
                // else
                // {
                //     cout << "NOT CONNECTED"<<endl;
                //      for (int j = 0; j < max_n; j++)
                //         cout<< vis[j];
                //     cout<<endl;
                // } 
                    
            }
            else
                cout << "HAS CYCLE"<<endl; 

            adj_map[edges[i][0]].insert ({edges[i][1]});
            adj_map[edges[i][1]].insert ({edges[i][0]});

           vis = all_not_vis;
        }
        
        return vector (0, 0);
        
    }
};
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