<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:
    void containsCycle(unordered_map<int, unordered_set<int>>& G,
                                vector<int>& V, int x, bool& b)
    {
        if (b == 1)
        {
            return;
        }
        for (int y : G[x])
        {
            if (V[y] == 1)
            {
                b = 1;
                return;
            }
            V[y] = 1;
            G[y].erase(x);
            containsCycle(G, V, y, b);
        }
    }

    vector<int> findRedundantConnection(const vector<vector<int>>& E)
    {
        unordered_map<int, unordered_set<int>> G;
        int m = E.size();
        int n = 0;
        for (int i = 0; i < m; ++i)
        {
            G[E[i][0]].insert(E[i][1]);
            G[E[i][1]].insert(E[i][0]);
            n = max(E[i][0], n);
            n = max(E[i][1], n);
        }
        unordered_map<int, unordered_set<int>> H = G;
        for (int i = n - 1; i >= 0; --i)
        {
            G[E[i][0]].erase(E[i][1]);
            G[E[i][1]].erase(E[i][0]);
            bool b = 0;
            vector<int> V(n + 1);
            for (int x = 1; x <= n; ++x)
            {
                if (V[x] == 0)
                {
                    V[x] = 1;
                    containsCycle(G, V, x, b);
                }
            }
            if (b == 0)
            {
                return E[i];
            }
            G = H;
        }
        return {-1, -1};
    }
};
29196