<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 find(int u, vector<int> &parent){
            while(1){
                if(parent[u]<0){
                    return u;
                }
                u=parent[u];
            }
        }
    
    void unite( int &comp ,int u , int v, vector<int> &parent){
          int p1=find(u, parent);
          int p2=find(v, parent);
          if(p1==p2) return;
          if(parent[p1]>parent[p2]){
            parent[p1]+=parent[p2];
            parent[p2]=p1;
          }
          else{
            parent[p2]+=parent[p1];
            parent[p1]=p2;
          }
          comp--;
    }
    int kruskal(int n, vector<vector<int>> &edges, int inc, int skip){
        vector<int> parent(n, -1);
        int sum=0;
        int comp=n;
        if(inc!=-1){
            sum+=edges[inc][0];
            unite(comp, edges[inc][1], edges[inc][2], parent);
        }
        for(int i=0; i<edges.size(); i++){
            if(i==skip) continue;
             int u=edges[i][1];
             int v=edges[i][2];
             int w=edges[i][0];
             if(find(u, parent)!=find(v, parent)){
                  sum+=w;
                  unite(comp, u, v, parent);
             }
        }
       return comp!=1?INT_MAX:sum;
       
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        for(int i=0; i<edges.size(); i++){
            int u=edges[i][0],  v=edges[i][1], w=edges[i][2];
            edges[i][0]=w;
            edges[i][1]=u;
            edges[i][2]=v;
            edges[i].push_back(i);
        }
        sort(edges.begin(), edges.end());
        int minwt=kruskal(n, edges, -1, -1);
        vector<vector<int>> ans(2, vector<int>());
        for(int i=0; i<edges.size(); i++){
            int skip=kruskal(n, edges, -1, i);
            int inc=kruskal(n, edges, i, -1);
            if(skip>minwt){
                ans[0].push_back(edges[i][3]);
            }
            else if(inc==minwt){
                ans[1].push_back(edges[i][3]);
            }
        }
        return ans;
    }
};
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