<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 DisjointSet:
    def __init__(self, n):
        self.node2root = {i:i for i in range(n)}
        self.node2rank = {i:0 for i in range(n)}
    
    def find(self, node):
        if self.node2root[node] != node:
            self.node2root[node] = self.find(self.node2root[node])
        return self.node2root[node]
    
    def union(self, node1, node2):
        root1 = self.find(node1)
        root2 = self.find(node2)
        if root1 == root2:
            return False
        rank1 = self.node2rank[root1]
        rank2 = self.node2rank[root2]
        if rank1 <= rank2:
            self.node2root[root1] = root2
            self.node2rank[root2] += (rank1 == rank2)
        else:
            self.node2root[root2] = root1
        return True

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        ds = DisjointSet(n)
        critical = []
        pseudo_critical = []

        Edge = namedtuple('Edge', ['ind', 'src', 'dst', 'weight'])
        sorted_edges = [Edge(i, *edge) for i, edge in enumerate(edges)]
        sorted_edges.sort(key=lambda e: e.weight)
        for weight, same_weight_edges in itertools.groupby(sorted_edges, key=lambda e: e.weight):
            graph = defaultdict(set)
            etype2edges = defaultdict(list)
            for edge_ind, s, d, w in same_weight_edges:
                root1 = ds.find(s)
                root2 = ds.find(d)
                if root1 != root2:  # criteria 1
                    graph[root1].add(root2)
                    graph[root2].add(root1)
                    etype = f'{min(root1, root2)}<->{max(root1, root2)}'
                    etype2edges[etype].append(edge_ind)
            for etype, interchangeable_edges in etype2edges.items():
                root1, root2 = map(int, etype.split('<->'))
                if len(interchangeable_edges) > 1 or self.edge_in_cycle(root1, root2, graph):  # criteria 2
                    for edge in interchangeable_edges:
                        pseudo_critical.append(edge)
                else:
                    critical.append(interchangeable_edges[0])
                ds.union(root1, root2)
        return [critical, pseudo_critical]

    def edge_in_cycle(self, src, dst, graph):
        graph[src].remove(dst)
        graph[dst].remove(src)
        still_has_path = self.find(src, dst, graph, set())
        graph[src].add(dst)
        graph[dst].add(src)
        return still_has_path

    def find(self, node, target, graph, seen):
        if node == target:
            return True
        if node in seen:
            return False
        seen.add(node)
        for neigh in graph[node]:
            if self.find(neigh, target, graph, seen):
                return True
        return False
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