<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 collections import namedtuple


Edge = namedtuple("edge", ["source", "dest", "weight"])


class UnionFind:
    def __init__(self, *args):
        self.parent = {arg: arg for arg in args}
        self.size = {arg: 1 for arg in args}
        self.number_of_subsets = len(args)

    def make_set(self, x):
        if x not in self.parent:
            self.parent[x] = x
            self.size[x] = 1
            self.number_of_subsets += 1

    def is_root(self, x):
        return self.parent[x] == x

    def different_subsets(self, x, y):
        return not self.same_subsets(x, y)

    def same_subsets(self, x, y):
        return self.find(x) == self.find(y)

    def find(self, x):
        if not self.is_root(x):
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        if self.different_subsets(x, y):
            x = self.find(x)
            y = self.find(y)
            if self.size[x] < self.size[y]:
                x, y = y, x
            self.parent[y] = x
            self.size[x] += self.size[y]
            self.number_of_subsets -= 1

    def is_single_subset(self):
        return self.number_of_subsets == 1


def kruskals(n, edges, forced = None, already_sorted = False):
    if not already_sorted:
        edges = sorted(edges, key = lambda edge : edge.weight)

    mst = UnionFind(*range(n))
    weight = 0

    if forced is not None:
        weight += forced.weight
        mst.union(forced.source, forced.dest)

    for e in edges:
        x = mst.find(e.source)
        y = mst.find(e.dest)
        if x != y:
            weight += e.weight
            mst.union(x, y)

    return weight


def critical_and_pseudocritical_edges(n, edges):
    edges = [Edge(*e) for e in edges]
    mst_weight = kruskals(n, edges)
    critical, pseudo_critical = [], []

    for i, e in enumerate(edges):
        new_edges = edges[:i] + edges[i + 1:]
        weight_without_edge = kruskals(n, new_edges)
        print(mst_weight, weight_without_edge, i)
        if weight_without_edge > mst_weight or weight_without_edge < mst_weight:
            critical.append(i)
        elif weight_without_edge == mst_weight:
            weight_with_edge = kruskals(n, new_edges, e)
            if weight_with_edge == mst_weight:
                pseudo_critical.append(i)

    result = []
    result.append(critical)
    result.append(pseudo_critical)
    return result


class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n, edges):
        return critical_and_pseudocritical_edges(n, edges)
1485