<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 UnionFind:
    def __init__(self, size:int)->None:
        self.size = size
        self.root = [i for i in range(size)]
        self.rank = [1]*size

    def find(self, x:int)->int:
        if x != self.root[x]:
            self.root[x] = self.find(self.root[x])
        return self.root[x]

    def union(self, x:int, y:int) -> bool:
        root_x, root_y = self.find(x), self.find(y)
        if root_x != root_y:
            if self.rank[root_x] > self.rank[root_y]:
                self.root[root_y] = root_x
            elif self.rank[root_x] < self.rank[root_y]:
                self.root[root_x] = root_y
            else:
                self.root[root_y] = root_x
                self.rank[root_x] +=1
            self.size-=1
            return True
        return False
    
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)]  #Add original indices of edge to edges array
        edges.sort(key=lambda x: x[2])  #Sort in increasing order of weight

        # O(E)
        def kruskal(n, edges, bannedIdx, includeIdx):
            uf = UnionFind(n)
            totalWeight = 0
            if includeIdx != -1:
                u, v, w, _ = edges[includeIdx]
                uf.union(u, v)
                totalWeight += w
                n -= 1

            for i, (u, v, weight, _) in enumerate(edges):
                if i == bannedIdx: continue
                if uf.union(u, v):
                    totalWeight += weight
                    n -= 1
                if n == 1: break  # A bit optimize since we found enough n-1 edges
            return math.inf if n > 1 else totalWeight

        criticalEdges = []
        mstWeight = kruskal(n, edges, -1, -1)
        for i in range(len(edges)):  # O(E * E)
            totalWeight = kruskal(n, edges, i, -1)
            if totalWeight > mstWeight:  # the mst increasing -> ith edge is critical
                criticalEdges.append(edges[i][3])  # add original index

        pseudoCriticalEdges = []
        for i in range(len(edges)):  # O(E * E)
            if edges[i][3] in criticalEdges: continue  # We only care non-critical edges
            totalWeight = kruskal(n, edges, -1, i)
            if totalWeight == mstWeight:  # the mst the same -> ith edge is pseudo-critical
                pseudoCriticalEdges.append(edges[i][3])  # add original index

        return [criticalEdges, pseudoCriticalEdges]
      
509