<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>

from typing import List
import heapq

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        def find(parent, x):
            if parent[x] != x:
                parent[x] = find(parent, parent[x])
            return parent[x]
        
        def union(parent, rank, x, y):
            rootX = find(parent, x)
            rootY = find(parent, y)
            if rootX != rootY:
                if rank[rootX] > rank[rootY]:
                    parent[rootY] = rootX
                elif rank[rootX] < rank[rootY]:
                    parent[rootX] = rootY
                else:
                    parent[rootY] = rootX
                    rank[rootX] += 1
        
        def kruskal(edges, n, skip_edge=None, force_edge=None):
            parent = list(range(n))
            rank = [0] * n
            total_weight = 0
            edge_count = 0
            
            if force_edge is not None:
                u, v, weight = force_edge
                union(parent, rank, u, v)
                total_weight += weight
                edge_count += 1
            
            for i, (u, v, weight) in enumerate(edges):
                if i == skip_edge:
                    continue
                if find(parent, u) != find(parent, v):
                    union(parent, rank, u, v)
                    total_weight += weight
                    edge_count += 1
            
            if edge_count == n - 1:
                return total_weight
            return float('inf')
        
        # Sort edges based on weight
        edges = sorted(enumerate(edges), key=lambda x: x[1][2])
        edge_list = [e[1] for e in edges]
        
        # Calculate the weight of the MST
        original_mst_weight = kruskal(edge_list, n)
        
        critical_edges = []
        pseudo_critical_edges = []
        
        for i, edge in enumerate(edge_list):
            # Check if edge is critical
            if kruskal(edge_list, n, skip_edge=i) > original_mst_weight:
                critical_edges.append(edges[i][0])
            else:
                # Check if edge is pseudo-critical
                if kruskal(edge_list, n, force_edge=edge) == original_mst_weight:
                    pseudo_critical_edges.append(edges[i][0])
        
        return [critical_edges, pseudo_critical_edges]

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