<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>

from typing import List
class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        edges = sorted([(w, u, v, i) for i , (u, v, w) in enumerate(edges)])
        def kruskal(n, edges, banned_edge = None, forced = None):
            parent = list(range(n))
            num = [0] * n

            def find(node):
                if parent[node] != node:
                    parent[node] = find(parent[node])
                return parent[node]
            
            def unionFind(u, v):
                rootU = find(u)
                rootV = find(v)
                if rootU != rootV:
                    if num[rootU] > num[rootV]:
                        parent[rootV] = rootU
                    elif num[rootU] < num[rootV]:
                        parent[rootU] = rootV
                    else:
                        parent[rootV] = rootU
                        num[rootU] += 1
            
            weight = 0
            edges_used = 0

            if forced is not None:
                _, u, v, _ = edges[forced]
                if find(u) != find(v):
                    unionFind(u, v)
                    weight += edges[forced][0]
                    edges_used += 1
            
            for i in range(len(edges)):
                if i == banned_edge:
                    continue
                w, u, v, _ = edges[i]
                if find(u) != find(v):
                    unionFind(u, v)
                    weight += w
                    edges_used += 1
                    if edges_used == n - 1:
                        break
            
            if edges_used == n - 1:
                return weight
            else:
                return float('inf')
        
        weight = kruskal(n, edges)

        critical = []
        pseudo_critical = []

        for i in range(len(edges)):
            if kruskal(n, edges, banned_edge = i) > weight:
                critical.append(edges[i][3])
            elif kruskal(n, edges, forced = i) == weight:
                pseudo_critical.append(edges[i][3])
        
        return[critical, pseudo_critical]
633