<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 add_e(unordered_map<int, vector<int>> &m,int u, int v) {
        if (m.count(u)) {
            m[u].push_back(v);
        } else {
            m.insert(make_pair(u, vector<int> {v}));
        }
    }
    bool Connected(unordered_map<int, vector<int>> m, int u, int v) {
        int st[1005] = {0};
        queue<int> q;
        st[u] = 1;
        q.push(u);
        while (!q.empty()) {
            int a = q.front();
            q.pop();
            vector<int> vv = m[a]; 
            for (int i = 0; i < vv.size(); i ++) {
                int b = vv[i];
                if (a == u && b == v || a == v && b == u || st[b] == 1) continue;
                if (b == v) return true;
                st[b] = 1;
                q.push(b);
            }
        }
        return false;
    }
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        unordered_map<int, vector<int>> m;
        for (int i = 0; i < edges.size(); i ++) {
            int u = edges[i][0];
            int v = edges[i][1];
            add_e(m, u, v);
            add_e(m, v, u);
        }
        for (int i = edges.size() - 1; i >= 0; i --) {
            int u = edges[i][0];
            int v = edges[i][1];
            if (Connected(m, u, v)) return edges[i];
        }
        return vector<int>{};
    }
};
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