<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>


template <typename T>
class DisjointSet
{
    unordered_map<T, T> parent;
    unordered_map<T, int> rank;
    public:
    int num_sets;
    DisjointSet(){
        num_sets=0;
    }
    void makeSet(vector<T> universe)
    {
        parent.clear();
        rank.clear();
        for (T i : universe)
        {
            parent[i] = i;
            rank[i] = 0;
        }
        num_sets = universe.size();
    }
    T find(T k)
    {
        if (parent[k] == k)
        {
            return k;
        }
        return parent[k] = find(parent[k]);
    }
    bool Union(T a, T b)
    {
        T pa = find(a);
        T pb = find(b);
        if (pa == pb)
        {
            return false;
        }
        if (rank[pa] > rank[pb])
        {
            parent[pb] = pa;
        }
        else if (rank[pa] < rank[pb])
        {
            parent[pa] = pb;
        }
        else
        {
            parent[pa] = pb;
            rank[pb]++;
        }
        this->num_sets = this->num_sets - 1;
        return true;
    }
};

bool comparator(pair<int, int> a, pair<int, int> b){
    if (a.first == b.first){
        return a.second > b.second;
    }
    return a.first < b.first;
}

class Solution {
public:
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        DisjointSet<int> ds1;
        vector<int> universe(n);
        iota(universe.begin(), universe.end(), 0);

        ds1.makeSet(universe);

        for (int i = 0; i < edges.size(); i++){
            edges[i].push_back(i);
        }

        sort(edges.begin(), edges.end(), [](vector<int> &a, vector<int> &b){
            return a[2] < b[2];
        });

        int min_cost = 0;
        for (int i = 0; i < edges.size(); i++){
            if (ds1.Union(edges[i][0], edges[i][1])){
                min_cost += edges[i][2];
            }
        }
        vector<vector<int>> res(2);
        DisjointSet<int> ds;
        for(int i = 0; i < edges.size(); i++){
            ds.makeSet(universe);
            int cost = 0;
            for (int j = 0; j < edges.size(); j++){
                if (i != j){
                    bool test = ds.Union(edges[j][0], edges[j][1]);
                    if (test){
                        cost += edges[j][2];
                    }
                }
            }
            if (ds.num_sets > 1 || cost > min_cost){
                res[0].push_back(edges[i][3]);
            }
            else{
                ds.makeSet(universe);
                ds.Union(edges[i][0], edges[i][1]);
                cost = edges[i][2];
                for (int j = 0; j < edges.size(); j++){
                    if (i != j){
                        if (ds.Union(edges[j][0], edges[j][1])){
                            cost += edges[j][2];
                        }
                    }
                }
                if (cost == min_cost){
                    res[1].push_back(edges[i][3]);
                }
            }
        }
        return res;
    }
};
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