<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 Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        class UnionFind:
            def __init__(self, n):
                self.parent = list(range(n))
                self.rank = [0] * n
            
            def find(self, u):
                if self.parent[u] != u:
                    self.parent[u] = self.find(self.parent[u])
                return self.parent[u]
            
            def union(self, u, v):
                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.parent[root_v] = root_u
                    elif self.rank[root_u] < self.rank[root_v]:
                        self.parent[root_u] = root_v
                    else:
                        self.parent[root_v] = root_u
                        self.rank[root_u] += 1
                    return True
                return False
        



        # Add index to edges for tracking original position
        for i in range(len(edges)):
            edges[i].append(i)
        
        # Sort edges by weight
        edges.sort(key=lambda x: x[2])
        
        # Function to calculate MST weight with optional skip/include edge
        def kruskal(skip_edge=-1, include_edge=-1):
            uf = UnionFind(n)
            weight = 0
            edges_used = 0
            
            if include_edge != -1:
                u, v, w, _ = edges[include_edge]
                uf.union(u, v)
                weight += w
                edges_used += 1
            
            for i in range(len(edges)):
                if i == skip_edge:
                    continue
                u, v, w, _ = edges[i]
                if uf.union(u, v):
                    weight += w
                    edges_used += 1
                if edges_used == n - 1:
                    break
            
            return weight if edges_used == n - 1 else float('inf')
        
        # Find the MST weight
        mst_weight = kruskal()
        
        critical = []
        pseudo_critical = []
        
        for i in range(len(edges)):
            # Check if it's a critical edge
            if kruskal(skip_edge=i) > mst_weight:
                critical.append(edges[i][3])
            # Check if it's a pseudo-critical edge
            elif kruskal(include_edge=i) == mst_weight:
                pseudo_critical.append(edges[i][3])
        
        return [critical, pseudo_critical]
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