<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 DisjointSet{
	vector<int> rank,parent,size;
public : 
      DisjointSet(int n){
      	size.resize(n+1,1);
      	rank.resize(n+1,0);
      	parent.resize(n+1);
      	for(int i=0;i<=n;i++){
      		parent[i]= i; 
      	}
      }
      int findUPar(int node){
      	if(node==parent[node])
      		return node;
      	return parent[node]=findUPar(parent[node]);
      }
      void unionBysize(int u,int v){
      	int ulp_u=findUPar(u);
      	int ulp_v=findUPar(v);
      	if(ulp_u==ulp_v)return;
      	if(size[ulp_u]<size[ulp_v]){
      		parent[ulp_u]=ulp_v;
      		size[ulp_v]+=size[ulp_u];
      	}
      	else {
      		parent[ulp_v]=ulp_u;
      		size[ulp_u]+=size[ulp_v];
      	}
      }
};
class Solution {
public:
int find(vector<pair<int, pair<int, pair<int,int>>>> &vp,int indx,int toadd,int u,int v,int n){
    DisjointSet ds(vp.size());
    int mst=0;
    for(auto it:vp){
        int u=it.second.second.first;
        int v=it.second.second.second;
        int dis=it.first;
        int edgeIndex=it.second.first;
        if(edgeIndex==indx)continue;
        int par_u=ds.findUPar(u);
        int par_v=ds.findUPar(v);
        if(par_u!=par_v){
            ds.unionBysize(u,v);
            mst+=dis;
        }
    }
    int cc = ds.findUPar(0);
        for(int i=1; i<n; ++i){
            if(ds.findUPar(i) != cc) return 1e9;
        }
        return mst;
    return mst;
}
int find1(vector<pair<int, pair<int, pair<int,int>>>> &vp,int indx,int toadd,int u,int v,int wt,int n){
    DisjointSet ds(n);
    int mst=wt;
    ds.unionBysize(u,v);
    for(auto it:vp){
        int u=it.second.second.first;
        int v=it.second.second.second;
        int dis=it.first;
        int edgeIndex=it.second.first;
       // if(edgeIndex==indx)continue;
        int par_u=ds.findUPar(u);
        int par_v=ds.findUPar(v);
        if(par_u!=par_v){
            ds.unionBysize(u,v);
            mst+=dis;
        }
    }
        return mst;
}
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<pair<int, pair<int, pair<int,int>>>> vp;
        for(int i = 0; i < edges.size(); i++){
            int dis = edges[i][2];
            int u = edges[i][0];
            int v = edges[i][1];
            vp.push_back({dis, {i, {u, v}}});
        }
        sort(vp.begin(),vp.end());
        DisjointSet ds(n);
        int mst=0;
        for(auto it:vp){
            int u=it.second.second.first;
            int v=it.second.second.second;
            int dis=it.first;
            int egdeIndex=it.second.first;
            int par_u=ds.findUPar(u);
            int par_v=ds.findUPar(v);
            if(par_u!=par_v){
                ds.unionBysize(u,v);
                mst+=dis;
            }
        }
        vector<int>critical;
        vector<int>pseudo_critical;
        unordered_set<int> critcalset;
        
        for(auto it:vp){
            int u=it.second.second.first;
            int v=it.second.second.second;
            int dis=it.first;
            int egdeIndex=it.second.first;
            int FindMst=find(vp,egdeIndex,-1,u,v,n);
            if(FindMst>mst){
                critcalset.insert(egdeIndex);
                critical.push_back(egdeIndex);
            }
        }
        for(auto it:vp){
            int u=it.second.second.first;
            int v=it.second.second.second;
            int dis=it.first;
            int egdeIndex=it.second.first;
            int FindMst=find1(vp,egdeIndex,-1,u,v,dis,n);
            if(FindMst==mst  && !critcalset.count(egdeIndex)){
                pseudo_critical.push_back(egdeIndex);
            }
        }
        return {critical,pseudo_critical};
    }
};
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