<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:
    class DisjointSet {
    vector<int> rank, parent, size;
public:
    DisjointSet(int n) {
        rank.resize(n + 1, 0);
        parent.resize(n + 1);
        size.resize(n + 1);
        for (int i = 0; i <= n; i++) {
            parent[i] = i;
            size[i] = 1;
        }
    }

    int findUPar(int node) {
        if (node == parent[node])
            return node;
        return parent[node] = findUPar(parent[node]);
    }

    void unionByRank(int u, int v) {
        int ulp_u = findUPar(u);
        int ulp_v = findUPar(v);
        if (ulp_u == ulp_v) return;
        if (rank[ulp_u] < rank[ulp_v]) {
            parent[ulp_u] = ulp_v;
        }
        else if (rank[ulp_v] < rank[ulp_u]) {
            parent[ulp_v] = ulp_u;
        }
        else {
            parent[ulp_v] = ulp_u;
            rank[ulp_u]++;
        }
    }
    
    void unionBySize(int u, int v) {
        int ulp_u = findUPar(u);
        int ulp_v = findUPar(v);
        if (ulp_u == ulp_v) return;
        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];
        }
    }
};    int spanningTree(int V, vector<vector<int>> &edges, int skip_edge, int add_edge)
    {
        // 1 - 2 wt = 5
        /// 1 - > (2, 5)
        // 2 -> (1, 5)

        // 5, 1, 2
        // 5, 2, 1
        
        DisjointSet ds(V);
        int mstWt = 0;
        if(add_edge!=-1){
                mstWt += edges[add_edge][2];
                ds.unionBySize(edges[add_edge][0], edges[add_edge][1]); 
        } 
        
        
        for (int i = 0; i<edges.size();i++) {
            if(skip_edge==i)continue;
            vector<int> vec = edges[i];
            int wt = vec[2];
            int u = vec[0];
            int v = vec[1];
            int idx = vec[3];
            if (ds.findUPar(u) != ds.findUPar(v)) {
                mstWt += wt;
                ds.unionBySize(u, v);
            }
              
        }
        for(int i=0;i<V;i++){
            if(ds.findUPar(i)!=ds.findUPar(0)){
                return INT_MAX;
            }
        }
        return mstWt;
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
         
      
         for(int i=0;i<edges.size();i++){
             edges[i].push_back(i);
        }
        auto lambda =[&](vector<int>& vec1, vector<int>&vec2){
              return vec1[2] < vec2[2];
        };
        sort(edges.begin(),edges.end(),lambda);

        int MST_wt = spanningTree(n,edges,-1,-1);
        vector<int> critical;
        vector<int> psuedoCritical;
        for(int i=0;i<edges.size();i++){
              
              int skipWt = spanningTree(n,edges,i,-1);
              int addWt = spanningTree(n,edges,-1,i);
              if(skipWt > MST_wt ){
                critical.push_back(edges[i][3]);
              }
              
              else if(addWt==MST_wt){
                    psuedoCritical.push_back(edges[i][3]);
              }

        }
        
        return {critical,psuedoCritical};
    }
};
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