<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 size=0; 
   static bool comp(vector<int>&v1,vector<int>&v2)   //comperator function
    {
        return v1[2]<v2[2];
    }
    
    int findpar(vector<int>&par,int u)   //find parent of a given nodes                      
    {
        if(u==par[u])return u;
        return par[u]=findpar(par,par[u]);
    }
	
    bool findunion(vector<int>&par,int u1,int v1,vector<int>&rank)
    {
        int u=findpar(par,u1),v=findpar(par,v1);
        if(u==v)return false;
        if(rank[u]<rank[v])
           par[u]=v;
        else if(rank[u]>rank[v])
           par[v]=u;
        else
        {
            par[v]=u;
            rank[u]++;
        }
        
         size--;  //this helps us to find disjoint set or (all the components of the graph)
        
       return true;
         
    }
    
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
          
        vector<int>null;
        for(int i = 0; i < edges.size(); i++){
            edges[i].push_back(i);                 //edges vector now become edges = [[0,1,1,0],[1,2,1,1],[2,3,2,2],[0,3,2,3],[0,4,3,4],[3,4,3,5],[1,4,6,6]] (this helps us to return the indices of edges)
          
        }
        
        sort(edges.begin(),edges.end(),comp);  //sort edges on the basis of its weight
    
        vector<vector<int>>ans(2);
        int original_cost=mst(n,edges,null,null);  //finding cost of original MST
        //cout<<original_cost<<"\n";
        for(auto it:edges)
        {  
           int exclude_cost=mst(n,edges,null,it);    //checking the cost is greater or not if we exclude this edge 
          
            if(original_cost<exclude_cost)
                ans[0].push_back(it[3]);
            else
            {
            
                int include_cost=mst(n,edges,it,null);   // check the cost is equal or not if we include this edge 
                 
               if(original_cost==include_cost)
                  ans[1].push_back(it[3]);   
            }    
        }
           return ans;
    }
    
    int mst(int n,vector<vector<int>>&edges,vector<int>&include,vector<int>& exclude)
    {
        vector<int>par(n),rank(n,0);
        for(int i=0;i<n;i++)
        {
            par[i]=i;
        }
        size=n;                       //size=n all the edges are disjoint set right now
        int cost=0;
        if(include.size())    //if include vector is not empty  then we will include that edge into out graph
        {
            findunion(par,include[0],include[1],rank);
            cost+=include[2];
           
        }
        
        for(auto it:edges)
        {   if(it==exclude)continue;   //if exclude vector is = to the "it(iterator)" then we'll  not include that edge and continue , else we'll include that edge 
            if(findunion(par,it[0],it[1],rank))
            {
                cost+=it[2];
            }
        }
       return size==1?cost:INT_MAX;   //here the size is = to the total no of disjoint components if the component is greter then 1 then return inf else return cost
    }    
};

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