Redundant Connection
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0684.Redundant%20Connection/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0684.Redundant%20Connection/images/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>
Depth-First Search; Breadth-First Search; Union Find; Graph
class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int n = edges.size();
        vector<int> p(n);
        iota(p.begin(), p.end(), 0);
        function<int(int)> find = [&](int x) {
            return x == p[x] ? x : p[x] = find(p[x]);
        };
        for (int i = 0;; ++i) {
            int pa = find(edges[i][0] - 1);
            int pb = find(edges[i][1] - 1);
            if (pa == pb) {
                return edges[i];
            }
            p[pa] = pb;
        }
    }
};
