<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 dsu
{
	vector<int>rank,parent,size;
	public:
		dsu(int n){
			rank.resize(n+1,0);
			parent.resize(n+1,0);
			size.resize(n+1,1);
			for(int i = 0; i <= n; i++){
				parent[i]=i;
			}
		}
		int findPar(int node){
			if(node==parent[node]){
				return node;
			}
			return parent[node]=findPar(parent[node]);
		}
		void unionBySize(int u, int v){
			int ultimate_parent_of_u = findPar(u);
			int ultimate_parent_of_v = findPar(v);
			if(ultimate_parent_of_v==ultimate_parent_of_u){
				return;
			}
			if(size[ultimate_parent_of_u]<=size[ultimate_parent_of_v]){
				parent[ultimate_parent_of_u] = ultimate_parent_of_v;
				size[ultimate_parent_of_v]+=size[ultimate_parent_of_u];
			}
			else{
				parent[ultimate_parent_of_v]=ultimate_parent_of_u;
				size[ultimate_parent_of_u]+=size[ultimate_parent_of_v];	
			}
		}
};
class Solution {
public:
    static bool cmp(vector<int>&a,vector<int>&b){
            return a[2]<b[2];
        }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        
         vector<vector<int>>nedge;
        int index = 0;
              int weight = 0;
        for(auto i:edges){
            vector<int>small = {i[2],i[0],i[1],index};
            nedge.push_back(small);
            index++;
        }
        sort(nedge.begin(),nedge.end());
        dsu ds1(n);

        for(auto e:nedge){
            // cout<<e[0]<<" "<<e[1]<<" "<<e[2]<<endl;
              if(ds1.findPar(e[1]) != ds1.findPar(e[2])){
                ds1.unionBySize(e[1],e[2]);
                weight += e[0];
            }
        }
      
        // weight is mst ka weight
           unordered_set<int>s;
        vector<vector<int>>ans(2);
        for(vector<int>leave : nedge){
            int cnt = 0;
            dsu ds(n);
            for(vector<int>edge:nedge){
                if(edge == leave){
                    continue;
                }
                if(ds.findPar(edge[1]) != ds.findPar(edge[2])){
                    ds.unionBySize(edge[1],edge[2]);
                    cnt += edge[0];
                }
            }
            if(cnt != weight){
                s.insert(leave[3]);// nhi loge to weight increase hoga
                ans[0].push_back(leave[3]);
            }
            // cout<<leave[3]<<" "<<cnt<<" "<<weight<<endl;
        }
          for(vector<int>take : nedge){
            if(s.find(take[3]) != s.end()){
                continue;
            }
            dsu ds(n);
            ds.unionBySize(take[1],take[2]);
            int cnt = take[0];
            for(vector<int>edge:nedge){
                if(edge == take){
                    continue;
                }
                if(ds.findPar(edge[1]) != ds.findPar(edge[2])){
                    ds.unionBySize(edge[2],edge[1]);
                    cnt += edge[0];
                }
            }
            if(cnt == weight){
                ans[1].push_back(take[3]);
            }
        }
        return ans;

    }
};
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