<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 collections import namedtuple
from math import inf

Edge = namedtuple("Edge", ["source", "dest", "weight"])

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

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX != rootY:
            if self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
            elif self.rank[rootX] < self.rank[rootY]:
                self.parent[rootX] = rootY
            else:
                self.parent[rootY] = rootX
                self.rank[rootX] += 1

def kruskals(n, edges, forced_edge=None, already_sorted=False):
    if not already_sorted:
        edges.sort(key=lambda edge: edge.weight)
    
    uf = UnionFind(n)
    mst_weight = 0
    mst_edges = 0

    # If we have a forced edge, add it first
    if forced_edge:
        uf.union(forced_edge.source, forced_edge.dest)
        mst_weight += forced_edge.weight
        mst_edges += 1
    
    for edge in edges:
        if uf.find(edge.source) != uf.find(edge.dest):
            uf.union(edge.source, edge.dest)
            mst_weight += edge.weight
            mst_edges += 1
        if mst_edges == n - 1:
            break
    
    # If we didn't use exactly n-1 edges, return infinity (disconnected graph)
    return mst_weight if mst_edges == n - 1 else inf

def critical_and_pseudocritical_edges(n, edges):
    # Re-index the edges to keep track of the original indices
    indexed_edges = [(i, Edge(*edge)) for i, edge in enumerate(edges)]
    
    # Find the weight of the original MST
    mst_weight = kruskals(n, [edge for _, edge in indexed_edges])
    
    critical, pseudo_critical = [], []
    
    for i, (idx, edge) in enumerate(indexed_edges):
        # Check if the edge is critical by removing it and computing the MST
        weight_without_edge = kruskals(n, [e for j, e in indexed_edges if j != i])
        if weight_without_edge > mst_weight:
            critical.append(idx)
        else:
            # Check if the edge is pseudo-critical by forcing its inclusion
            weight_with_forced_edge = kruskals(n, [e for j, e in indexed_edges if j != i], forced_edge=edge)
            if weight_with_forced_edge == mst_weight:
                pseudo_critical.append(idx)
    
    return [critical, pseudo_critical]

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n, edges):
        return critical_and_pseudocritical_edges(n, edges)

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