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

        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

    # Indent kruskal to be a method of Solution
    def kruskal(self, n, edges, exclude_edge=-1, include_edge=-1):
        uf = self.UnionFind(n)  # Use self.UnionFind
        mst_weight = 0
        edges_used = 0

        # Include the forced edge first if specified
        if include_edge != -1:
            u, v, w, _ = edges[include_edge]
            if uf.union(u, v):
                mst_weight += w
                edges_used += 1

        for i, (u, v, w, _) in enumerate(edges):
            if i == exclude_edge:
                continue
            if uf.union(u, v):
                mst_weight += w
                edges_used += 1
            if edges_used == n - 1:
                break

        # If we didn't use enough edges, return a large number to indicate failure
        return mst_weight if edges_used == n - 1 else float('inf')

    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        # Add the original index to the edges for easier identification later
        edges = [(u, v, w, i) for i, (u, v, w) in enumerate(edges)]
        edges.sort(key=lambda x: x[2])  # Sort by weight

        # Step 1: Find the weight of the original MST
        original_mst_weight = self.kruskal(n, edges)

        critical_edges = []
        pseudo_critical_edges = []

        for i in range(len(edges)):
            u, v, w, idx = edges[i]
            
            # Step 2: Check if it's a critical edge by excluding it
            if self.kruskal(n, edges, exclude_edge=i) > original_mst_weight:  # Call with self.kruskal
                critical_edges.append(idx)
            else:
                # Step 3: Check if it's a pseudo-critical edge by including it
                if self.kruskal(n, edges, include_edge=i) == original_mst_weight:  # Call with self.kruskal
                    pseudo_critical_edges.append(idx)

        return [critical_edges, pseudo_critical_edges]

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