<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 {
public:
int findparent(int u,vector<int>&parent)
{
    if(u==parent[u])
      return u;
    return parent[u]=findparent(parent[u],parent);
}
bool union1(int u,int v,vector<int>&parent,vector<int>&rank)
{
   int pu=findparent(u,parent);
   int pv=findparent(v,parent);
   if(pu==pv)
     return false;
   else if(rank[pu]>=rank[pv])
   {
    parent[pv]=pu;
    rank[pu]+=rank[pv];
   }
   else
   {
    parent[pu]=pv;
    rank[pv]+=rank[pu];
   }
   return true;

}
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<vector<int>>alledges;
        vector<int>parent(n,-1),rank(n,0);
        unordered_map<int,int>um;
        for(int i=0;i<n;i++)
        {
            parent[i]=i;
        }
        for(int i=0;i<edges.size();i++)
        {
            alledges.push_back({edges[i][2],edges[i][0],edges[i][1],i});

        }
        sort(alledges.begin(),alledges.end());
        int sum=0;
        for(auto i:alledges)
        {
            int u=i[1];
            int v=i[2];
            int w=i[0];
            int index=i[3];
            if(union1(u,v,parent,rank)){
                sum+=w;
                um[index]++;
            }
        }
        vector<int>cric,uncric;
      for(int i=0;i<edges.size();i++)
      { int wt=0,incwt=0,flag=0;
        int comp=n;
        vector<int>parent2(n,-1),parent3(n,-1),rank2(n,0),rank3(n,0);
        for(int i=0;i<n;i++)
        {
            parent2[i]=i;
            parent3[i]=i;
        }
        for(auto temp:alledges)
        {
            int u=temp[1];
            int v=temp[2];
            int w=temp[0];
            int index=temp[3];
        if(i==index)
        {
            union1(u,v,parent3,rank3);
                incwt+=w;
        }
        }    
        for(auto j:alledges)
        {
            int u=j[1];
            int v=j[2];
            int w=j[0];
            int index=j[3];
            if(index!=i)
            {
            if(union1(u,v,parent2,rank2)){
                wt+=w;
                comp--;
            }
            }
           if(union1(u,v,parent3,rank3))
           {
             incwt+=w;
           }
           
        }
        if(sum<wt || sum>wt && comp!=1)
         cric.push_back(i);
        else if(incwt==sum && wt==sum)
        {
         uncric.push_back(i);
        }
      }
      return {cric,uncric};
    }
};
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