<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 find(self,u,parent):
        if u == parent[u] : return u
        return self.find(parent[u], parent)

    def unionDSU(self,u,v,parent) :
        p1 = self.find(u, parent)
        p2 = self.find(v, parent)
        parent[p2] = p1

    def mst(self,n,edges,includeEdge,excludeEdge) :

        parent=[]
        for i in range(n): 
            parent.append(i);
        
        ans = 0
        count = 0

        if len(includeEdge) != 0 :
            self.unionDSU(includeEdge[0], includeEdge[1], parent)
            ans += includeEdge[2]
            count += 1
        
        for edge in edges:
            u = edge[0];
            v = edge[1];
            cost = edge[2];

            if len(excludeEdge) != 0 and excludeEdge[0] == u and excludeEdge[1] == v and excludeEdge[2] == cost :
                continue
            

            if len(includeEdge) != 0 and includeEdge[0] == u and includeEdge[1] == v and includeEdge[2] == cost :
                continue
            

            p1 = self.find(u, parent)
            p2 = self.find(v, parent)

            if p1 != p2:
                self.unionDSU(p1, p2, parent)
                ans += cost
                count += 1
            
        
        return  ans if count==n-1 else float('inf')
    
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        
        originalEdges=[]
        for edge in edges : 
            originalEdge=[edge[0], edge[1], edge[2]]
            originalEdges.append(originalEdge)
        

        result=[]
        criticalEdges=[]
        pseudoCriticalEdges=[]

        
        edges=sorted(edges, key = lambda x: x[2])

        emptyVector=[]
        originalCost = self.mst(n, edges, emptyVector, emptyVector)

        
        for i in range(len(originalEdges)): 

            excludedCost = self.mst(n, edges, emptyVector, originalEdges[i])
            includedCost = self.mst(n, edges, originalEdges[i], emptyVector)

            if excludedCost > originalCost :
                criticalEdges.append(i)
            elif includedCost == originalCost : 
                pseudoCriticalEdges.append(i)
            
        
        result.append(criticalEdges)
        result.append(pseudoCriticalEdges)
        return result
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