<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]]:
        hm = {}
        for i, e in enumerate(edges):
            hm[tuple(e)] = i
        sortedEdges = sorted(edges, key = lambda edge: edge[2])
        uf = UF()
        minWeight = self.kruskals(n, sortedEdges, uf)
        res = [[], []]
        for i, e in enumerate(sortedEdges):
            edge = sortedEdges.pop(i)
            uf = UF()
            temp = self.kruskals(n, sortedEdges, uf)
            if temp > minWeight:
                res[0].append(hm[tuple(edge)])
            else:
                uf = UF()
                uf.insert(e[0])
                uf.insert(e[1])
                uf.union(e[0], e[1])
                temp = 0
                temp += e[2]
                temp = self.kruskals(n, sortedEdges, uf, temp)
                if temp == minWeight:
                    res[1].append(hm[tuple(edge)])
            sortedEdges.insert(i, edge)
        return res

    def kruskals(self, n, sortedEdges, uf, minWeight = 0):
        for e in sortedEdges:
            if uf.check(e[0]) and uf.check(e[1]):
                if uf.find(e[0], e[1]):
                    continue
                else:
                    uf.union(e[0], e[1])
                    minWeight += e[2]
            else:
                uf.insert(e[0])
                uf.insert(e[1])
                uf.union(e[0], e[1])
                minWeight += e[2]
        return minWeight if uf.isComplete(n) else float('inf')

class UF:
    def __init__(self):
        self.parent = {}
        self.size = {}
        self.maxSize = 0

    def isComplete(self, n):
        return self.maxSize == n

    def check(self, p):
        return p in self.parent

    def insert(self, p):
        if self.check(p):
            return
        self.parent[p] = -1
        self.size[p] = 1
        self.maxSize = max(self.maxSize, 1)

    def find(self, p, q):
        while self.parent[p] != -1:
            p = self.parent[p]
        while self.parent[q] != -1:
            q = self.parent[q]
        return p == q

    def union(self, p, q):
        while self.parent[p] != -1:
            p = self.parent[p]
        while self.parent[q] != -1:
            q = self.parent[q]
        if self.size[p] > self.size[q]:
            self.parent[q] = p
            self.size[p] += self.size[q]
            self.maxSize = max(self.maxSize, self.size[p])
        else:
            self.parent[p] = q
            self.size[q] += self.size[p]
            self.maxSize = max(self.maxSize, self.size[q])
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