<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:
    # import copy
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        # sets = []
        # for i, j in edges:
        #     if i in x and j in x:
        #         return [i, j]
        #     elif i in x and j not in x:
        #         x.append(j)
        #     elif i not in x and j in x:
        #         x.append(i)
        #     else:
        #         x.append(i)
        #         x.append(j)
        # d = {}
        # for i in range(len(edges)):
        #     sets.append(set([i + 1]))
        #     d[i + 1] = set([i + 1])
        # print(sets)
        # for i, j in edges:
        #     # i -= 1
        #     # j -= 1 
        #     if d[i] == d[j] :
        #         # print(sets[i],sets[j])
        #         return [i, j]
        #     else:
        #         # print(sets[i],sets[j])
        #         superset = d[i].union(d[j])
        #         d[i] = superset
        #         d[j] = superset
        #         # sets[i] = sets[i].union(sets[j])
        #         # sets[i] = copy.copy(setx)
        #         # sets[i].update(sets[j])
        #         # sets[j] = copy.deepcopy(sets[i])
        #         # sets[j] = sets[i].deepcopy()
        #         # sets[j] =  copy.copy(sets[i])
        #     print(d)
        par = [i for i in range(len(edges) + 1)]
        # rank = [1] * (len(edges) + 1)
        def find(n):
            p = par[n]
            while p != par[p]:
                par[p] = par[par[p]]
                p = par[p]
            return p
        def union(n1, n2):
            p1, p2 = find(n1), find(n2)

            if p1 == p2:
                return False
            par[p1] = p2
            # if rank[p1] > rank[p2]:
            #     par[p2] = p1
            #     rank[p1] += rank[p2]
            # else:
            #     par[p1] = p2
            #     rank[p2] += rank[p1]
            return True

        for n1, n2 in edges:
            if not union(n1, n2):
                return [n1, n2]
        
        
16900