<p>Given a weighted undirected connected graph with <code>n</code>&nbsp;vertices numbered from <code>0</code> to <code>n - 1</code>,&nbsp;and an array <code>edges</code>&nbsp;where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>, weight<sub>i</sub>]</code> represents a bidirectional and weighted edge between nodes&nbsp;<code>a<sub>i</sub></code>&nbsp;and <code>b<sub>i</sub></code>. A minimum spanning tree (MST) is a subset of the graph&#39;s edges that connects all vertices without cycles&nbsp;and with the minimum possible total edge weight.</p>

<p>Find <em>all the critical and pseudo-critical edges in the given graph&#39;s minimum spanning tree (MST)</em>. An MST edge whose deletion from the graph would cause the MST weight to increase is called a&nbsp;<em>critical edge</em>. On&nbsp;the other hand, a pseudo-critical edge is that which can appear in some MSTs but not all.</p>

<p>Note that you can return the indices of the edges in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/ex1.png" style="width: 259px; height: 262px;" /></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[0,1,1],[1,2,1],[2,3,2],[0,3,2],[0,4,3],[3,4,3],[1,4,6]]
<strong>Output:</strong> [[0,1],[2,3,4,5]]
<strong>Explanation:</strong> The figure above describes the graph.
The following figure shows all the possible MSTs:
<img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/msts.png" style="width: 540px; height: 553px;" />
Notice that the two edges 0 and 1 appear in all MSTs, therefore they are critical edges, so we return them in the first list of the output.
The edges 2, 3, 4, and 5 are only part of some MSTs, therefore they are considered pseudo-critical edges. We add them to the second list of the output.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/ex2.png" style="width: 247px; height: 253px;" /></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[0,1,1],[1,2,1],[2,3,1],[0,3,1]]
<strong>Output:</strong> [[],[0,1,2,3]]
<strong>Explanation:</strong> We can observe that since all 4 edges have equal weight, choosing any 3 edges from the given 4 will yield an MST. Therefore all 4 edges are pseudo-critical.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= edges.length &lt;= min(200, n * (n - 1) / 2)</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt; n</code></li>
	<li><code>1 &lt;= weight<sub>i</sub>&nbsp;&lt;= 1000</code></li>
	<li>All pairs <code>(a<sub>i</sub>, b<sub>i</sub>)</code> are <strong>distinct</strong>.</li>
</ul>

class UnionFind:
    def __init__(self, n):
        self.root = list(range(n))
        self.rank = [1] * n

    def find(self, u) -> int:
        if self.root[u] != u:
            self.root[u] = self.find(self.root[u])
        return self.root[u]

    def union(self, u, v) -> bool:
        root_u = self.find(u)
        root_v = self.find(v)

        if root_u != root_v:
            if self.rank[root_u] > self.rank[root_v]:
                self.root[root_v] = root_u
            elif self.rank[root_u] < self.rank[root_v]:
                self.root[root_u] = root_v
            else:
                self.root[root_v] = root_u
                self.rank[root_u] += 1
            return True
        return False

class Solution:
    def findMST(self, n, edges, include=None, exclude=None):
        uf = UnionFind(n)   
        cost = 0
        edge_count = 0

        if include is not None:
            uf.union(edges[include][0], edges[include][1])
            cost += edges[include][2]
            edge_count += 1

        for i, (u, v, w, _) in enumerate(edges):
            if i == exclude:
                continue
            
            if uf.union(u, v):
                cost += w
                edge_count += 1
            
            if edge_count == n - 1:
                break

        return cost if edge_count == n - 1 else float('inf')

    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        sorted_edges = sorted([[u, v, w, i] for i, [u, v, w] in enumerate(edges)], key=lambda x: x[2])
        
        min_weights = self.findMST(n, sorted_edges)

        critical = []
        non_critical = []

        for i in range(len(sorted_edges)):
            if self.findMST(n, sorted_edges, exclude=i) > min_weights:
                critical.append(sorted_edges[i][3])
            elif self.findMST(n, sorted_edges, include=i) == min_weights:
                non_critical.append(sorted_edges[i][3])
        
        return [critical, non_critical]
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