<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]:
        MxN = 1001
        link = [x for x in range(MxN)]
        size = [1 for x in range(MxN)]
        # size = [x for x in range(MxN)] # we're going to do this without balancing at the start
        # the basis of union find is assigning a representative for each set
        # at first, each node has it's own representative, then to connect two nodes together
        # you assign the representative of one to the other, then balance, we won't balance here
        def find(x: int) -> int:
            if link[x] == x:
                return x
            return find(link[x])

        def union(a: int, b: int) -> bool:
            parent_a, parent_b = find(a), find(b)
            if parent_a == parent_b: return True
            if size[parent_a] < size[parent_b]: 
                parent_a, parent_b = parent_b, parent_a
            size[a] += size[b]
            link[parent_b] = parent_a # making parent_a the representative of b
            return False
        # def display_parents() -> None:
        #     nodes = set([u for u, _ in edges] + [v for _, v in edges])
            # print({x: link[x] for x in nodes})
        
        res = None
        for u, v in edges:
            if union(u, v): res = [u, v]
            # display_parents()
        return res
        
        
17500