<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.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

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n, edges):
        # Add index to edges for tracking
        for i, edge in enumerate(edges):
            edge.append(i)
        
        # Sort edges by weight
        edges.sort(key=lambda x: x[2])
        
        # Find MST weight
        def find_mst_weight(include_edge=-1, exclude_edge=-1):
            uf = UnionFind(n)
            weight = 0
            if include_edge != -1:
                uf.union(edges[include_edge][0], edges[include_edge][1])
                weight += edges[include_edge][2]
            
            for i, (u, v, w, _) in enumerate(edges):
                if i == exclude_edge:
                    continue
                if uf.find(u) != uf.find(v):
                    uf.union(u, v)
                    weight += w
            
            # Check if all nodes are connected
            root = uf.find(0)
            for i in range(1, n):
                if uf.find(i) != root:
                    return float('inf')
            
            return weight
        
        # Find MST weight without any edge excluded
        base_weight = find_mst_weight()
        
        critical, pseudo_critical = [], []
        
        for i, (u, v, w, idx) in enumerate(edges):
            # Check if the edge is critical
            if find_mst_weight(exclude_edge=i) > base_weight:
                critical.append(idx)
            # Check if the edge is pseudo-critical
            elif find_mst_weight(include_edge=i) == base_weight:
                pseudo_critical.append(idx)
        
        return [critical, pseudo_critical]

# Example usage:
solution = Solution()
n = 5
edges = [[0,1,1],[1,2,1],[2,3,2],[0,3,2],[0,4,3],[3,4,3],[1,4,6]]
print(solution.findCriticalAndPseudoCriticalEdges(n, edges))  # Output: [[0,1], [2,3,4,5]]
841