<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{
public:
    unordered_map<int,int>p,sz;
    int find(int u){
        if(!p.count(u))p[u]=u,sz[u]=1;
        return u==p[u]?u:(p[u]=find(p[u]));
    }
    bool unite(int x,int y){
        x = find(x);
        y = find(y);
        if(x==y)return false;
        if(sz[x] >sz[y])swap(x,y);
        sz[y]+=sz[x];
        p[x]=p[y];
        return true;
    }
};
class Solution {
public:
    int getMSTCost(vector<vector<int>>& edges, int skipIdx, int n, int include){
        DSU dsu;
        int cost=0;
        if(include!=-1){
           if(dsu.unite(edges[include][0],edges[include][1])){
                cost += edges[include][2];
            } 
        }
        for(int idx=0;idx<edges.size();idx++){
            if(skipIdx == idx || include == idx)continue;
            if(dsu.unite(edges[idx][0],edges[idx][1])){
                cost += edges[idx][2];
            }
        }
        int comp=0;
        for(int i=0;i<n;i++)comp+=(dsu.find(i) == i);
        if(comp>1)return INT_MAX;
        return cost;
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& E) {
        vector<vector<int>>res(2);
        vector<vector<int>>edges;
        int ind = 0;
        for(auto &i:E){
            edges.push_back(i);
            edges.back().push_back(ind++);
        }
        sort(edges.begin(),edges.end(),[&](vector<int>&a,vector<int>&b){
            return a[2] < b[2];
        });
        
        int cost = getMSTCost(edges,-1,n,-1);
        for(int idx=0;idx<edges.size();idx++){
            int newCost = getMSTCost(edges,idx,n,-1);
            if(cost < newCost){
                res[0].push_back(edges[idx][3]);
            }else {
                newCost = getMSTCost(edges,-1,n,idx);
                if(newCost == cost)res[1].push_back(edges[idx][3]);
            }
        }
        return res;
    }
};
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