<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:
    int n,set_size, *parent, *rank;
    UnionFind(){}
    UnionFind(int a) { //a: set size, {1,2...n}
        n=set_size=a;
        parent=new int[n];
        rank=new int[n];
        for (int i=0;i<n;i++) parent[i]=i, rank[i]=1;
    }

    int find(int x) { //O(A(n))
        if (x!=parent[x]) return parent[x]=find(parent[x]); //path-compression
        return x;
    }

    bool merge(int x, int y) { //O(A(n))
        int xroot=find(x), yroot=find(y);
        if (xroot!=yroot) { //rank-compression
            if (rank[xroot]>=rank[yroot]) {
                parent[yroot]=xroot;
                rank[xroot]+=rank[yroot];
            } else {
                parent[xroot]=yroot;
                rank[yroot]+=rank[xroot];
            }
            set_size--;
            return 1;
        }
        return 0;
    }

    void reset() {
        set_size=n;
        for (int i=0;i<n;i++) parent[i]=i, rank[i]=1;  
    }
};

int get_mst_cost(int n, vector<vector<int>> &edges, int unmark, unordered_set<int> &marks) { //O(E)
    UnionFind uf(n);
    int mst_cost=0;
    for (int i=0;i<edges.size();i++) {
        if (i==unmark) continue;
        auto &v=edges[i];
        int &x=v[0], &y=v[1];

        int k=i+1;
        while (k<edges.size() && edges[k][2]==edges[i][2]) {
            if (uf.find(edges[k][0])!=uf.find(edges[k][1]))
                marks.insert(edges[k][3]);
            k++;
        }

        if (uf.find(x)!=uf.find(y)) {
            mst_cost+=v[2];
            marks.insert(edges[i][3]);
            uf.merge(x,y);
        }
    }
    return uf.set_size==1?mst_cost:1e9;
}
class Solution {
public:
vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
    for (int i=0;i<edges.size();i++) edges[i].push_back(i);
    sort(edges.begin(), edges.end(), [](vector<int> &v1, vector<int> &v2) {
        return v1[2]<v2[2];
    }); //O(ElogE)

    unordered_set<int> marks, __, crit;

    int orig_cost=get_mst_cost(n,edges,-1,marks);
    for (int i=0;i<edges.size();i++) { //O(E^2)
        int wt=get_mst_cost(n,edges,i,__);
        if (wt>orig_cost) crit.insert(edges[i][3]);
    }
    vector<int> Psuedo; for (auto &_:marks) if (crit.find(_)==crit.end()) 
        Psuedo.push_back(_);
    vector<int> Crit;for (auto &_:crit) Crit.push_back(_);
    return {Crit, Psuedo};
}
};
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