<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 findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        parent = [i for i in range(n)]
        def find(x):
            if x == parent[x]:
                return x
            parent[x] = find(parent[x])
            return parent[x]
        def union(x, y):
            xp, yp = find(x), find(y)
            parent[xp] = yp
        
        weight = {}
        edge2idx = {}
        for idx, edge in enumerate(edges):
            weight.setdefault(edge[2], []).append(edge[:2])
            edge2idx[tuple(edge[:2])] = idx
        pseudo_critical = set()
        critical = set()
        def dfs(node, cur_depth):
            if depth[node]: return depth[node]
            min_depth = depth[node] = cur_depth
            for neighbor in tedges[node]:
                if depth[neighbor] == cur_depth - 1:
                    continue
                res = dfs(neighbor, cur_depth+1)
                if res <= cur_depth:
                    pseudo_critical.add(tedge2idx[node, neighbor])
                elif tedge2idx[(node, neighbor)] not in pseudo_critical:
                    critical.add(tedge2idx[node, neighbor])
                min_depth = min(min_depth, res)
            return min_depth
        for w in sorted(weight.keys()):
            tedges = {}
            tedge2idx = {}
            nodes = set()
            for src, dst in weight[w]:
                find(src)
                find(dst)
                if parent[src] == parent[dst]: continue
                if parent[src] in tedges.get(parent[dst], {}):
                    pseudo_critical.add(edge2idx[src, dst])
                    pseudo_critical.add(tedge2idx[parent[src], parent[dst]])
                    continue
                tedges.setdefault(parent[src], set()).add(parent[dst])
                tedges.setdefault(parent[dst], set()).add(parent[src])
                tedge2idx[parent[src], parent[dst]] = tedge2idx[parent[dst], parent[src]] = edge2idx[src, dst]
                nodes |= {parent[src], parent[dst]}
            #print(tedges, tedge2idx)
            depth = {node:0 for node in nodes}
            for node in nodes:
                dfs(node, 2)
            for src in tedges:
                for dst in tedges[src]:
                    if find(src) != find(dst):
                        union(src, dst)
        return critical, pseudo_critical
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