<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, n):
        self.par = {}
        self.rank = {}

        for i in range(0, n):
            self.par[i] = i
            self.rank[i] = 1
    
    # Find parent of n, with path compression.
    def find(self, n):
        p = self.par[n]
        while p != self.par[p]:
            self.par[p] = self.par[self.par[p]]
            p = self.par[p]
        return p

    # Union by height / rank.
    # Return false if already connected, true otherwise.
    def union(self, n1, n2):
        p1, p2 = self.find(n1), self.find(n2)
        if p1 == p2:
            return False
        
        if self.rank[p1] > self.rank[p2]:
            self.par[p2] = p1
            self.rank[p1] = self.rank[p1] + self.rank[p2]
        else:
            self.par[p1] = p2
            self.rank[p2] = self.rank[p2] + self.rank[p1]
        return True

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:

        # Step 1: Find Minimum Spanning Tree Weight
        uf = UnionFind(n)
        minH = []
        for a, b, w in edges:
            heapq.heappush(minH, [w, a, b])
        span_weight = 0
        while minH:
            w, a, b = heapq.heappop(minH)
            if uf.union(a, b):
                span_weight += w

        
        # Step 2 & 3: Get Critical and Pseudo Critical edges...Brute force approach
        indexEdges = []
        index = 0

        print(span_weight, indexEdges)
        for a, b, w in edges:
            indexEdges.append([w, a, b, index])
            index += 1

        critical = []
        pseudo = []
        indexEdges.sort(key=lambda w: w[0])

        for w, a, b, index in indexEdges:
            # Get Critical edges: Without a and b edge
            cWeight = 0
            cUF = UnionFind(n)
            for cW, cA, cB, cIndex in indexEdges:
                if index != cIndex and cUF.union(cA, cB):
                    cWeight += cW
            if (max(cUF.rank) != n - 1) or cWeight != span_weight:
                critical.append(index)
                continue

            # Get Pseudo edges: With a and b edge
            pWeight = w
            pUF = UnionFind(n)
            pUF.union(a, b) # forcefully add
            print("Inside before", a, b, index, w)
            for pW, pA, pB, pIndex in indexEdges:
                if pUF.union(pA, pB):
                    print("Inside edge", pA, pB)
                    pWeight += pW
            if pWeight == span_weight:
                print("Inside hi", pWeight, pUF.par, pUF.rank, pseudo)
                pseudo.append(index)
    
        return [critical, pseudo]
800