<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:
    class disjoint{
     public:
     vector<int> parent,rank;
     disjoint(int n)
     {
         parent.resize(n);
            rank.resize(n, 1);
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
     }
      int findpar(int u)
      {
        if(parent[u]==u) return u;
        return parent[u]=findpar(parent[u]);
      }

      void unionbyrank(int u,int v)
      {
        int x=findpar(u);
        int y=findpar(v);
        if(x==y) return;
        if(rank[x]<rank[y])
        {
             parent[x]=y;
        }
        else if(rank[y]<rank[x])
        {
             parent[y]=x;
        }
        else
        {
            parent[x]=y;
            rank[y]++;
        }
      }
    };
    int calmst(int n, vector<vector<int>>& edges,int flag,vector<int>& temp)
    {
        //flag=0 //add
        //flag=1 //skip
        disjoint ds(n);
        int wt=0;
        if(flag==0)
        {
            ds.unionbyrank(temp[0],temp[1]);
            wt+=temp[2];
        }
  

       
        for(int i=0;i<edges.size();i++)
        {
         int a=edges[i][0];
         int b=edges[i][1];
         int c=edges[i][2];
         if (flag == 1 && edges[i] == temp) continue;
         if(ds.findpar(a)!=ds.findpar(b) )
         {
               ds.unionbyrank(a,b);
               wt+=c;
         }
        }
 
        int root = ds.findpar(0);
        for (int i = 1; i < n; i++) {
            if (ds.findpar(i) != root) {
                return INT_MAX; 
            }
        }
        return wt;
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
      vector<vector<int>> adj;
        vector<vector<int>> ans(2);

        for (int i = 0; i < edges.size(); i++) {
            vector<int> temp = edges[i];
            temp.push_back(i);
            adj.push_back(temp);
        }
            sort(adj.begin(), adj.end(), [&](vector<int> a, vector<int> b){
    return (a[2] == b[2]) ? a[3] < b[3] : a[2] < b[2];
});
      vector<int> temp5;
        int mstOfTree = calmst(n, adj, -1, temp5);
        vector<int> criticalEdges;
        vector<int> pseudoEdges;

        for (int i = 0; i < adj.size(); i++) {
            vector<int> temp = adj[i];
            int mst = calmst(n, adj, 1, temp);
            if (mst > mstOfTree) {
                criticalEdges.push_back(temp[3]);
            }
            else 
            {
                 int mst1= calmst(n, adj, 0, temp);
                 if(mst1==mstOfTree){
                pseudoEdges.push_back(temp[3]);
                 }
            }
        }


        

        set<int> s(criticalEdges.begin(), criticalEdges.end());
        vector<int> temp(s.begin(), s.end());
        ans[0] = temp;

        set<int> s2(pseudoEdges.begin(), pseudoEdges.end());
        vector<int> temp2(s2.begin(), s2.end());
        ans[1] = temp2;

        return ans;
    }
};
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