<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 DisjointSet{
    vector<int> parent,size;
public:
    DisjointSet(int n){
        parent.resize(n+1);
        size.resize(n+1,1);
        for(int i=0;i<=n;i++){
            parent[i] = i;
        }
    }
    int findP(int i){
        if(parent[i]==i){
            return i;
        }
        return parent[i] = findP(parent[i]);
    }
    bool unionBySize(int u,int v){
        int ulp_u = findP(u);
        int ulp_v = findP(v);
        if(ulp_u == ulp_v)return false;
        if(size[ulp_u] < size[ulp_v]){
            parent[ulp_u] = ulp_v;
            size[ulp_v] += size[ulp_u];
        }
        else{
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v];
        }
        return true;
    }
};

class Solution {
public:
    void dfs(int node,int parent,int timer,vector<int> &discTime,vector<int> &lowTime,
    unordered_map<int,list<pair<int,int>>> &adj,unordered_map<int,bool> &vis,vector<int> &result){
        vis[node] = 1;
        discTime[node] = lowTime[node] = timer++;
        for(auto [adjNode,index]:adj[node]){
            if(adjNode==parent)continue;
            if(vis[adjNode]){
                lowTime[node] = min(lowTime[node],discTime[adjNode]);
            }
            else{
                dfs(adjNode,node,timer,discTime,lowTime,adj,vis,result);
                lowTime[node] = min(lowTime[node],lowTime[adjNode]);
                if(lowTime[adjNode] > discTime[node]){
                    result.push_back(index);
                }
            }
        }

    }
    vector<int> criticalConnections(int n, vector<vector<int>>& edges) {
        unordered_map<int,list<pair<int,int>>> adj;
        for(int i=0;i<edges.size();i++){
            int u = edges[i][0];
            int v = edges[i][1];
            adj[u].push_back({v,i});
            adj[v].push_back({u,i});
        }
        int timer = 0;
        vector<int> discTime(n,-1);
        vector<int> lowTime(n,-1);
        int parent = -1;
        unordered_map<int,bool> vis;
        vector<int> result;
        for(int i=0;i<n;i++){
            if(!vis[i]){
                dfs(i,parent,timer,discTime,lowTime,adj,vis,result);
            }
        }
        return result;
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<vector<int>> ans(2);
        vector<int> tm = criticalConnections(n,edges);
        unordered_map<int,bool> um;
        for(auto &it:tm){
            ans[0].push_back(it);
            um[it]=1;
        }
        DisjointSet ds(n);
        vector<pair<int,int>> temp;
        for(int i=0;i<edges.size();i++){
            temp.push_back({edges[i][2],i});
        }
        sort(temp.begin(),temp.end());
        int weight = 0;
        for(int i=0;i<temp.size();i++){
            if(ds.unionBySize(edges[temp[i].second][0],edges[temp[i].second][1])){
                weight += temp[i].first;
            }
        }
        for(int i=0;i<edges.size();i++){
            if(um[i])continue;
            DisjointSet sample(n);
            int w = 0;
            for(int j=0;j<temp.size();j++){
                if(temp[j].second==i)continue;
                if(sample.unionBySize(edges[temp[j].second][0],edges[temp[j].second][1])){
                    w += temp[j].first;
                }
            }
            if(w>weight){
                ans[0].push_back(i);
                um[i]=1;
            }
            else{
                DisjointSet sample2(n);
                w = edges[i][2];
                bool pp = sample2.unionBySize(edges[i][0],edges[i][1]);
                for(int j=0;j<temp.size();j++){
                    if(sample2.unionBySize(edges[temp[j].second][0],edges[temp[j].second][1])){
                        w += temp[j].first;
                    }
                }
                if(w>weight){
                    um[i]=1;
                }
            }
        }
        for(int i=0;i<edges.size();i++){
            if(!um[i])ans[1].push_back(i);
        }
        sort(ans[0].begin(),ans[0].end());
        return ans;
    }
};
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