<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]]:
        original_edges=edges[:]
        edges.sort(key=lambda x: x[2])
        
        def find(x,parent,rank):
            if parent[x]==x: return x
            parent[x]=find(parent[x],parent,rank)
            return parent[x]
        def union(x,y,parent,rank):
            x_rep=find(x,parent,rank)
            y_rep=find(y,parent,rank)
            if x_rep==y_rep: return True
            xrank=rank[x_rep]
            yrank=rank[y_rep]
            if xrank>yrank:
                parent[y_rep]=x_rep
            elif yrank>xrank:
                parent[x_rep]=y_rep
            else:
                parent[x_rep]=y_rep
                rank[y_rep]+=1
            return False
        def kruskal(e,flag):
            res=0
            count=0
            parent=[i for i in range(n)]
            rank=[0 for i in range(n)]
            if flag:
                u,v,w=original_edges[e]
                union(u,v,parent,rank)
                res=w
                count+=1
            for i in range(len(edges)):
                if e==-1 or edges[i]!=original_edges[e]:
                    u,v,w=edges[i]
                    if union(u,v,parent,rank)==False:
                        res+=w
                        count+=1
                        if count==n-1: break
            if count==n-1 : return (True,res)
            return (False,-1)
        res=kruskal(-1,False)

        critical=[]
        pseudocritical=[]
        for i in range(len(edges)):
            x=kruskal(i,False)
            if  x[0]==False or x[1]>res[1]:
                critical.append(i)
            else:
                x=kruskal(i,True)
                if x[1]==res[1]:
                    pseudocritical.append(i)
        
        return [critical,pseudocritical]

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