<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 UnionFindSet:
    def __init__(self, n=0):
        self.parents = {}
        self.ranks = {}
        self.count = 0
        for i in range(n):
            self.add(i)

    def add(self, p):
        self.parents[p] = p
        self.ranks[p] = 1
        self.count += 1

    def find(self, u):
        if u != self.parents[u]:
            self.parents[u] = self.find(self.parents[u])
        return self.parents[u]

    def union(self, u, v):
        pu, pv = self.find(u), self.find(v)
        if pu == pv: 
            return False
        if self.ranks[pu] < self.ranks[pv]:
            self.parents[pu] = pv
        elif self.ranks[pu] > self.ranks[pv]:
            self.parents[pv] = pu
        else:        
            self.parents[pv] = pu
            self.ranks[pu] += 1
        self.count -= 1
        return True

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        edges = [(u, v, w, i) for i, (u, v, w) in enumerate(edges)]
        edges.sort(key=lambda x: x[2])
        
        def find_mst_without_this_edge(edge_idx):
            union_find_set = UnionFindSet(n)
            ans = 0
            for i, (u, v, w, _) in enumerate(edges):
                if i == edge_idx:
                    continue
                if union_find_set.union(u, v):
                    ans += w
            parent = union_find_set.find(0)
            return ans if all(union_find_set.find(i) == parent for i in range(n)) else inf
        
        def find_mst_with_this_edge(edge_idx):
            union_find_set = UnionFindSet(n)
            u0, v0, w0, _ = edges[edge_idx]
            ans = w0
            union_find_set.union(u0, v0)
            for i, (u, v, w, _) in enumerate(edges):
                if i == edge_idx:
                    continue
                if union_find_set.union(u, v):
                    ans += w
            parent = union_find_set.find(0)
            return ans if all(union_find_set.find(i) == parent for i in range(n)) else inf
        
        base = find_mst_without_this_edge(-1)
        cri, p_cri = set(), set()
        for i in range(len(edges)):
            wgt_excl = find_mst_without_this_edge(i)
            if wgt_excl > base:
                cri.add(edges[i][3])
            else:
                wgt_incl = find_mst_with_this_edge(i)
                if wgt_incl == base:
                    p_cri.add(edges[i][3])
    
        return [cri, p_cri]
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