<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:
    def __init__(self):
        self.parent = []
        self.rank = []

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

    def union(self, node1, node2):
        root1 = self.find(node1)
        root2 = self.find(node2)

        if root1 != root2:
            if self.rank[root1] > self.rank[root2]:
                self.parent[root2] = root1
            elif self.rank[root1] < self.rank[root2]:
                self.parent[root1] = root2
            else:
                self.parent[root2] = root1
                self.rank[root1] += 1

    def kruskal(self, n, edges, exclude_edge=None, force_edge=None):
        """ Helper function to run Kruskal's algorithm, with the option to exclude or force an edge """
        self.parent = [i for i in range(n)]
        self.rank = [0 for _ in range(n)]
        weight = 0
        count = 0

        # Force an edge into the MST if provided
        if force_edge is not None:
            u, v, w, idx = edges[force_edge]
            self.union(u, v)
            weight += w
            count += 1

        # Iterate through the remaining edges and build the MST
        for idx, (u, v, w, original_idx) in enumerate(edges):
            if idx == exclude_edge:
                continue
            if self.find(u) != self.find(v):
                self.union(u, v)
                weight += w
                count += 1
            if count == n - 1:
                break

        # If the MST is complete, return the weight, otherwise return infinity
        return weight if count == n - 1 else float('inf')

    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        # Add index to each edge to track the original position
        for i in range(len(edges)):
            edges[i].append(i)  # Append the original index

        # Sort edges by weight
        edges.sort(key=lambda x: x[2])

        # Find the weight of the MST without removing or forcing any edges
        mst_weight = self.kruskal(n, edges)

        critical = []
        pseudo_critical = []

        for i in range(len(edges)):
            # Check if removing edge i increases the MST weight
            if self.kruskal(n, edges, exclude_edge=i) > mst_weight:
                critical.append(edges[i][3])  # Use original index
            # Check if forcing edge i into the MST doesn't change the weight
            elif self.kruskal(n, edges, force_edge=i) == mst_weight:
                pseudo_critical.append(edges[i][3])  # Use original index

        return [critical, pseudo_critical]

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