<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 = [0] * 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
            return True
        return False

def kruskals(n, edges, forced_edge=None, ignore_edge=None):
    uf = UnionFind(n)
    mst_weight = 0
    mst_edges = 0

    # If we have a forced edge, include it in the MST 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 edge == ignore_edge:
            continue
        if uf.union(edge.source, edge.dest):
            mst_weight += edge.weight
            mst_edges += 1
        if mst_edges == n - 1:
            break
    
    # If we did not include n-1 edges, then we didn't form a spanning tree
    return mst_weight if mst_edges == n - 1 else inf

def critical_and_pseudocritical_edges(n, edges):
    # Convert edges to a named tuple for better readability
    edges = [Edge(*e) for e in edges]

    # Sort edges based on weight
    edges = sorted(enumerate(edges), key=lambda x: x[1].weight)
    
    # Find the MST weight with no restrictions
    base_mst_weight = kruskals(n, [edge[1] for edge in edges])

    critical, pseudo_critical = [], []

    for i, (index, edge) in enumerate(edges):
        # Check if it's a critical edge
        if kruskals(n, [edge[1] for j, edge in enumerate(edges) if j != i]) > base_mst_weight:
            critical.append(index)
        # Check if it's a pseudo-critical edge
        elif kruskals(n, [edge[1] for edge in edges], forced_edge=edge) == base_mst_weight:
            pseudo_critical.append(index)

    return [critical, pseudo_critical]

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: list) -> list:
        return critical_and_pseudocritical_edges(n, edges)

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