<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:

    vector<int>parent;
    void disjoint(int size){
        parent.resize(size+1);
        for(int i=0;i<=size;i++)
        parent[i]=(i);
    }
    int find(int u){
        if(parent[u]==u)return u;
        return parent[u] = find(parent[u]);
    }
    void merge(int u,int v){
        int ua = find(u);
        int ub = find(v);
        parent[ua] = ub;
    }
    int help1(vector<vector<int>>& e,int j,int n){
        disjoint(n+1);
        vector<pair<int,pair<int,int>>>v;
        for(int i=0;i<e.size();i++){
            if(i==j)continue;
            v.push_back({e[i][2],{e[i][0],e[i][1]}});
        }
        sort(v.begin(),v.end());
        int mst_weight = 0;
        int edges = 0;
        for(int i=0;i<v.size();i++){
            auto x = v[i];
            int u = find(x.second.first);
            int v = find(x.second.second);
            if(u!=v){
                edges++;
                mst_weight += x.first;
                merge(u,v);
            } 
        }
        if(edges!=n-1)return INT_MAX;
        return mst_weight;
    }
    int help2(vector<vector<int>>& e,int j,int n){
        disjoint(n+1);
        int mst_weight = e[j][2];
        merge(e[j][1],e[j][0]);
        vector<pair<int,pair<int,int>>>v;
        for(int i=0;i<e.size();i++){
            if(i==j)continue;
            v.push_back({e[i][2],{e[i][0],e[i][1]}});
        }
        sort(v.begin(),v.end());
        for(int i=0;i<v.size();i++){
            auto x = v[i];
            int u = find(x.second.first);
            int v = find(x.second.second);
            if(u!=v){
                mst_weight += x.first;
                merge(u,v);
            } 
        }
        return mst_weight;
    }

    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        disjoint(n+1);
        int mst_weight = help1(edges,-1,n);
        vector<vector<int>>ans;
        vector<int>v1,v2;
        for(int i=0;i<edges.size();i++){
            int new_weight1 = help1(edges,i,n);
            int new_weight2 = help2(edges,i,n);
            if(new_weight1 > mst_weight){
                v1.push_back(i);
            }
           else if(new_weight2 == mst_weight){
                v2.push_back(i);
            }
        }
        ans.push_back(v1);
        ans.push_back(v2);
        return ans;
    }
};
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