<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 DSU{
        public:
        vector<int> parent;
        vector<int> subtreeSize;

        void init(int n){
            parent.resize(n);
            subtreeSize.assign(n, 1);
            for(int i = 0; i < n; i++) parent[i] = i;
        }

        int findPar(int n){
            if(parent[n] == n) return n;
            return parent[n] = findPar(parent[n]);
        }

        void merge(int n1, int n2){
            int p1 = findPar(n1);
            int p2 = findPar(n2);
            if(p1 == p2) return;
            if(subtreeSize[p1] < subtreeSize[p2]) swap(p1, p2);
            parent[p2] = p1;
            subtreeSize[p1] += subtreeSize[p2];
        
        }
    };

    static bool edgeWeightSort(vector<int> &edge1, vector<int> &edge2){
        return edge1[2] < edge2[2];
    }

    bool edgeIndexSort(vector<int> &edge1, vector<int> &edge2){
        return edge1[3] < edge2[3];
    }

    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        int m = edges.size();
        for(int i = 0; i < m; i++) edges[i].push_back(i);
        sort(edges.begin(), edges.end(), edgeWeightSort);    
        set<int> criticalEdges;
        set<int> pseudoCriticalEdges;
        int minSpanningTree = 0;
        DSU dsu;
        dsu.init(n);
        set<int> maintreeEdges;
        for(auto &edge: edges){
            int p1 = dsu.findPar(edge[0]), p2 = dsu.findPar(edge[1]);
            if(p1 == p2) continue;
            maintreeEdges.insert(edge[3]);
            dsu.merge(p1, p2);
            minSpanningTree += edge[2];
        }

        for(int i = 0; i < m; i++){
            // skipping i^th
            dsu.init(n);
            int currSpanningTree = 0;
            set<int> currtreeEdges;
            for(auto &edge: edges){
                if(edge[3] == i) continue;
                int p1 = dsu.findPar(edge[0]), p2 = dsu.findPar(edge[1]);
                if(p1 == p2) continue;
                currtreeEdges.insert(edge[3]);
                dsu.merge(p1, p2);
                currSpanningTree += edge[2];
            } 
            if(currSpanningTree > minSpanningTree || currtreeEdges.size() < n-1) criticalEdges.insert(i);
        }
        for(int i = 0; i < m; i++){
            // taking i^th
            dsu.init(n);
            int currSpanningTree = 0;
            int p1 = dsu.findPar(edges[i][0]), p2 = dsu.findPar(edges[i][1]);
            dsu.merge(p1, p2);
            currSpanningTree += edges[i][2];
            for(auto &edge: edges){
                int p1 = dsu.findPar(edge[0]), p2 = dsu.findPar(edge[1]);
                if(p1 == p2) continue;
                dsu.merge(p1, p2);
                currSpanningTree += edge[2];
            } 
            if(currSpanningTree == minSpanningTree && criticalEdges.find(edges[i][3]) == criticalEdges.end()) pseudoCriticalEdges.insert(edges[i][3]);
        }
        vector<int> criticalEdgeList;
        for(auto edgeIdx: criticalEdges) criticalEdgeList.push_back(edgeIdx);
        vector<int> pseudoCriticalEdgeList;
        for(auto edgeIdx: pseudoCriticalEdges) pseudoCriticalEdgeList.push_back(edgeIdx);

        vector<vector<int>> finalOutput = {criticalEdgeList, pseudoCriticalEdgeList};
        return finalOutput;
    }
};
47690