<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{
public:
    vector<int>rank,parent,size;
    DisjointSet(int n){
        rank.resize(n,0);
        parent.resize(n);
        size.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 unionByRank(int u,int v){
        int ulp_u=findUPar(u),ulp_v=findUPar(v);
        if(ulp_u==ulp_v)return;
        if(rank[ulp_u]<rank[ulp_v]){
            parent[ulp_u]=ulp_v;
        }
        else if(rank[ulp_v]<rank[ulp_u]){
            parent[ulp_v]=ulp_u;
        }
        else{
            parent[ulp_v]=ulp_u;
            rank[ulp_u]++;
        }
    }

    void unionBySize(int u,int v){
        int ulp_u=findUPar(u),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 krushkal(int n,vector<vector<int>>edg,int skip,int force){
        int sum=0;
        DisjointSet ds(n);
        if(force!=-1){
            ds.unionBySize(edg[force][1],edg[force][2]);
            sum+=edg[force][0];
        }
        
        for(int i=0;i<edg.size();i++){
            if(i!=skip && ds.findUPar(edg[i][1])!=ds.findUPar(edg[i][2])){
                ds.unionBySize(edg[i][1],edg[i][2]);
                sum+=edg[i][0];
            }
        }
        for(int i=0;i<n;i++){
            if(ds.findUPar(0)!=ds.findUPar(i))return INT_MAX;
        }
        return sum;
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<vector<int>>edg;
        for(int i=0;i<edges.size();i++){
            int u=edges[i][0],v=edges[i][1],wt=edges[i][2];
            edg.push_back({wt,u,v,i});
        }
        sort(edg.begin(),edg.end());
        vector<int>critical,pseudo;
        int mst=krushkal(n,edg,-1,-1);
        for(int i=0;i<edg.size();i++){
            if(krushkal(n,edg,i,-1)>mst)critical.push_back(edg[i][3]);
            else if(krushkal(n,edg,-1,i)==mst)pseudo.push_back(edg[i][3]);
        }
        return {critical,pseudo};
    }
};
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