<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 typing import List

class UnionFind:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.rank = [0] * n

    def find(self, x):
        if x != self.parent[x]:
            self.parent[x] = self.find(self.parent[x])  # Path compression
        return self.parent[x]

    def union(self, x, y):
        x_parent = self.find(x)
        y_parent = self.find(y)

        if x_parent == y_parent:
            return

        if self.rank[x_parent] > self.rank[y_parent]:
            self.parent[y_parent] = x_parent
        elif self.rank[x_parent] < self.rank[y_parent]:
            self.parent[x_parent] = y_parent
        else:
            self.parent[x_parent] = y_parent
            self.rank[y_parent] += 1
class Solution:
    def __init__(self):
        self.N=0
    def krushkal(self, edges, skip_edge, force_edge):
        sum_weight=0
        uf=UnionFind(self.N)
        if force_edge != -1:
            u,v,wt,idx=edges[force_edge]
            uf.union(u,v)
            sum_weight+=wt
        for i in range(len(edges)):
            if i==skip_edge:
                continue
            u,v,wt,idx=edges[i]
            if uf.find(u)!=uf.find(v):#Before adding an edge to the MST, you need to check if the two nodes it connects are already in the same set (i.e., connected).
                uf.union(u,v)
                sum_weight+=wt
        for i in range(self.N):
            if uf.find(i)!=uf.find(0):
                return float('inf')
        return sum_weight

    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        # self.N=n
        # for i in range(len(edges)):
        #     edges[i].append(i)
        # edges.sort(key=lambda x:x[2])
        # MST_WEIGHT=self.krushkal(edges,-1,-1)
        # critical=[]
        # pseudo_critical=[]
        # for i in range(len(edges)):
        #     if self.krushkal(edges, i, -1)>MST_WEIGHT:
        #         critical.append(edges[i][3])
        #     if self.krushkal(edges,-1,i)==MST_WEIGHT:
        #         pseudo_critical.append(edges[i][3])
        # return [critical, pseudo_critical]
        self.N = n
        for i in range(len(edges)):
            edges[i].append(i)  # Append original index
        edges.sort(key=lambda x: x[2])
        MST_WEIGHT = self.krushkal(edges, -1, -1)
        critical = []
        pseudo_critical = []
        for i in range(len(edges)):
            if self.krushkal(edges, i, -1) > MST_WEIGHT:
                critical.append(edges[i][3])
            elif self.krushkal(edges, -1, i) == MST_WEIGHT:
                pseudo_critical.append(edges[i][3])
        return [critical, pseudo_critical]

        
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