<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 heapq import heappush, heappop


class UnionFind:
    def __init__(self, n):
        self.parents = []
        self.sizes = []
        for i in range(n):
            self.parents.append(i)
            self.sizes.append(1)

    def find(self, x):
        p = x
        while p != self.parents[p]:
            self.parents[p] = self.parents[self.parents[p]]
            p = self.parents[p]
        return p

    def union(self, x, y):
        px = self.find(x)
        py = self.find(y)
        if px == py:
            return False
        if self.sizes[px] > self.sizes[py]:
            self.sizes[px] += self.sizes[py]
            self.parents[py] = px
        else:
            self.sizes[py] += self.sizes[px]
            self.parents[px] = py
        return True

def kruskal(uf, n, min_heap):
    n_edges = 0
    mst = 0
    while min_heap and n_edges < n - 1:
        w, a, b = heappop(min_heap)
        if uf.union(a, b):
            n_edges += 1
            mst += w
    return mst

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        min_heap = []
        for a, b, w in edges:
            heappush(min_heap, [w, a, b])
        
        uf = UnionFind(n)
        mst = kruskal(uf, n, min_heap)

        critical = []
        noncritical = []
        for i in range(len(edges)):
            min_heap = []
            for j in range(len(edges)):
                if i == j:
                    continue
                a, b, w = edges[j]
                heappush(min_heap, [w, a, b])
            uf = UnionFind(n)
            new_mst = kruskal(uf, n, min_heap)
            if new_mst != mst:
                critical.append(i)
            else:
                noncritical.append(i)

        pseudocritical = []
        for i in noncritical:
            a, b, w = edges[i]
            uf = UnionFind(n)
            uf.union(a, b)
            new_mst = w

            min_heap = []
            for a, b, w in edges:
                heappush(min_heap, [w, a, b])

            new_mst += kruskal(uf, n, min_heap)
            if new_mst == mst:
                pseudocritical.append(i)
        return [critical, pseudocritical]

966