<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 UF:
    def __init__(self, n):
        self.parents = list(range(n))
        self.ranks = [1] * (n+1)
    
    def find(self, a):
        if a != self.parents[a]:
            self.parents[a] = self.find(self.parents[a])
        return self.parents[a]
    
    def union(self, a, b):
        p1, p2 = self.find(a), self.find(b)
        if p1 != p2:
            if self.ranks[p1] > self.ranks[p2]:
                self.parents[p2] = p1
                self.ranks[p1] += self.ranks[p2]
            else:
                self.parents[p1] = p2
                self.ranks[p2] += self.ranks[p1]
            return False
        else:
            return True

class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        ret = None
        n = len(edges)
        uf = UF(n+1)
        for e in edges:
            print(e, uf.parents)
            if uf.union(e[0], e[1]):
                return e
        return ret
17400