<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 Disjoint{
    private:
    vector<int> rank,parent;
    public:
    Disjoint(int n){
        rank.resize(n,0);
        parent.resize(n);
        for(int i=0;i<n;i++){
            parent[i]=i;
        }
    }
    int ulp(int node){
        if(node==parent[node]){
            return node;
        }
        return parent[node]=ulp(parent[node]);
    }
    void unionByRank(int u,int v){
        int ulp_u=ulp(u);
        int ulp_v=ulp(v);
        if(ulp_u==ulp_v)return;
        if(rank[ulp_u]>rank[ulp_v]){
            parent[ulp_v]=ulp_u;
        }
        else if(rank[ulp_u]<rank[ulp_v]){
            parent[ulp_u]=ulp_v;
        }
        else{
            parent[ulp_v]=ulp_u;
            rank[ulp_u]++;
        }
    }
};



class Solution {
public:
    static bool compare(vector<int>&a,vector<int>&b){
        return a[2]<b[2];
    }
    int critical(int n,vector<vector<int>> &edges,vector<int> edge){
        Disjoint ds(n);
        int pu=edge[0];
        int pv=edge[1];
        int ans=0;
        for(int i=0;i<edges.size();i++){
            int u=edges[i][0];
            int v=edges[i][1];
            int wt=edges[i][2];
            if(u==pu&&v==pv){
                continue;//this is edges not considered
            }
            if(ds.ulp(u)==ds.ulp(v)){
                continue;
            }
            ans+=wt;
            ds.unionByRank(u,v);
        }
        for(int i=0;i<n;i++){
            if(ds.ulp(0)!=ds.ulp(i)){
                ans=1e8;
                break;
            }
        }
        return ans;
    }
    int pseudoCritical(int n,vector<vector<int>>& edges,vector<int> &arr){
        Disjoint ds(n);
        int ans=0;
        int pu=arr[0];
        int pv=arr[1];
        int pw=arr[2];
        ans+=pw;
        ds.unionByRank(pu,pv);
        for(int i=0;i<edges.size();++i){
            int u=edges[i][0];
            int v=edges[i][1];
            int w=edges[i][2];
            if(ds.ulp(u)==ds.ulp(v))
            continue;
            ans+=w;
            ds.unionByRank(u,v); 
        }
        return ans;
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<vector<int>> temp=edges;
        sort(temp.begin(),temp.end(),compare);
        int mstWt=mst(n,temp);
        vector<vector<int>> ans(2,vector<int>());
        for(int i=0;i<edges.size();i++){
            int newMst=critical(n,temp,edges[i]);
            if(newMst>mstWt){
                ans[0].push_back(i);
                continue;
            }
            newMst=pseudoCritical(n,temp,edges[i]);
            if(newMst==mstWt){
                ans[1].push_back(i);
            }

        }
        return ans;



        
    }
    int mst(int n,vector<vector<int>>&edges){//sorted edges here
        Disjoint ds(n);
        int ans=0;
        for(int i=0;i<edges.size();i++){
            int u=edges[i][0];
            int v=edges[i][1];
            int wt=edges[i][2];
            if(ds.ulp(u)==ds.ulp(v)){
                continue;
            }
            ans+=wt;
            ds.unionByRank(u,v);
        }
        return ans;
    }
};
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