<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:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        
        parent = {v: v for v in range(1, len(edges) + 1)}  # Each node starts as its own parent
        rank = {v: 1 for v in range(1, len(edges) + 1)}    # Each node starts with rank 1

        # ---------------------------------
        
        def find(u):
            # Path compression
            if parent[u] != u:
                parent[u] = find(parent[u])
            return parent[u]
        
        # ---------------------------------
        
        def union(u, v):
            parent_of_u = find(u)
            parent_of_v = find(v)
            
            if parent_of_u == parent_of_v:
                # u and v have the same root, meaning adding this edge forms a cycle
                return True
            
            # Union by rank: attach the tree with the smaller rank under the one with the larger rank
            if rank[parent_of_u] > rank[parent_of_v]:
                parent[parent_of_v] = parent_of_u
            elif rank[parent_of_u] < rank[parent_of_v]:
                parent[parent_of_u] = parent_of_v
            else:
                parent[parent_of_v] = parent_of_u
                rank[parent_of_u] += 1
            
            return False
        
        # ---------------------------------
        
        redundant = None
        for u, v in edges:
            if union(u, v):
                # If adding the edge (u, v) forms a cycle, it's the redundant edge
                redundant = [u, v]
                return redundant

17000