<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.pars = [i for i in range(n)]
        self.rank = [1 for i in range(n)]

    def find(self, n):
        p = self.pars[n]
        while p != self.pars[p]:
            self.pars[p] = self.pars[self.pars[p]]
            p = self.pars[p]
        return p

    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.pars[p2] = p1
            self.rank[p1] += self.rank[p2]
        else:
            self.pars[p1] = p2
            self.rank[p2] += self.rank[p1]
        return True

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

        for i, e in enumerate(edges):
            e.append(i)     # [a, b, weight, idx_original]

        edges.sort(key=lambda x: x[2])

        # determine base weight
        baseweight = 0
        uf = UnionFind(n)
        for e in edges:
            if uf.union(e[0], e[1]):
                baseweight += e[2]

        # determine critical and pseudocritical edges
        critical, pseudocritical = [], []
        for i in range(len(edges)):
            idx = edges[i][3]

            # critical edge check
            uf = UnionFind(n)
            weight = 0
            for e in edges:
                if e[3] != idx:
                    if uf.union(e[0], e[1]):
                        weight += e[2]

            if max(uf.rank) < n or weight > baseweight:
                critical.append(idx)
                continue

            # pseudocritical edge check
            uf = UnionFind(n)
            uf.union(edges[i][0], edges[i][1])
            weight = edges[i][2]
            for e in edges:
                if uf.union(e[0], e[1]):
                    weight += e[2]
            
            if weight == baseweight:
                pseudocritical.append(idx)


        return [critical, pseudocritical]




        
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