<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 UnionFind{
    public:
    vector<int> parent;
    UnionFind(int n){
        parent.resize(n);
        for(int i=0;i<n;i++){
            parent[i]=i;
        }
    }
    int find(int p){
        if(parent[p]==p)
            return p;
        return parent[p]=find(parent[p]);
    }
    bool merge(int x,int y){
        int px=find(x);
        int py=find(y);
        if(px!=py){
            parent[py]=px;
            return true;
        }
        return false;
    }
};

    

class Solution {
public:
int calculate(vector<vector<int>>& edges,int escape,int n,int incl){
        UnionFind uf(n);
        int temp=0;
        if(incl!=-1){
            temp+=edges[incl][2];
            uf.merge(edges[incl][0],edges[incl][1]);
        }
        int i=0;
        for(auto v:edges){
            if(i==escape){i++;continue;}
            if(uf.find(v[0])!=uf.find(v[1])){
                temp+=v[2];
                uf.merge(v[0],v[1]);
            }
            i++;
        }
        for(int j=1;j<n;j++){
            if(uf.find(0)!=uf.find(j))
                return INT_MAX;
        }
        return temp;

    }
    static bool comp(vector<int> &a,vector<int> &b){
        return a[2]<b[2];
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        for(int i=0;i<edges.size();i++){
            edges[i].push_back(i);
        }
        vector<int> criti;
        vector<int> pcriti;
        sort(edges.begin(),edges.end(),comp);
        int base=calculate(edges,-1,n,-1);
        for(int i=0;i<edges.size();i++){
            if(calculate(edges,i,n,-1)>base){
                criti.push_back(edges[i][3]);
            }
            else if(calculate(edges,-1,n,i)==base){
                pcriti.push_back(edges[i][3]);
            }
        }
        vector<vector<int>> ans;ans.push_back(criti);ans.push_back(pcriti);
        return ans;

    }
};
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