<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> par, sz;
            DSU(int n) {
                par.resize(n);
                for(int i=0; i<n; i++) par[i]=i;
                sz.resize(n, 1);
            }

            int findPar(int node) {
                if(node==par[node]) return node;
                return par[node] = findPar(par[node]);
            }

            void un(int u, int v) {
                int up = findPar(u), vp = findPar(v);
                if(up==vp) return;

                if(sz[up]>vp) {
                    sz[up]+=sz[vp];
                    par[vp]=up;
                } else {
                    sz[vp]+=sz[up];
                    par[up]=vp;
                }
            }
    };

    int tc = 0;

    int f(int n, vector<vector<int>>& edges, int skip) {
        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;

        DSU* dsu = new DSU(n);

        int m = edges.size(), ind=0;
        for(auto &i: edges) {
            if(ind!=skip) {
                pq.push({i[2], {i[0], i[1]}});
            }
            ind++;
        }

        int cost = 0, ed=0;

        while(!pq.empty()) {
            auto it = pq.top();
            int u = it.second.first, v = it.second.second, w = it.first;

            if(dsu->findPar(u)!=dsu->findPar(v)) {
                ed++;
                dsu->un(u, v);
                cost+=w;
            }

            pq.pop();
        }

        if(ed!=n-1) return INT_MAX;
        return cost;
    }

    int f2(int n, vector<vector<int>>& edges, int skip) {
        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;

        DSU* dsu = new DSU(n);

        int cost = edges[skip][2];
        dsu->un(edges[skip][0], edges[skip][1]);

        int m = edges.size(), ind=0;
        for(auto &i: edges) {
            if(ind!=skip) {
                pq.push({i[2], {i[0], i[1]}});
            }
            ind++;
        }

        while(!pq.empty()) {
            auto it = pq.top();
            int u = it.second.first, v = it.second.second, w = it.first;

            if(dsu->findPar(u)!=dsu->findPar(v)) {
                dsu->un(u, v);
                cost+=w;
            }

            pq.pop();
        }

        return cost;
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        tc = 0;
        unordered_map<int, int> st;
        int mini = f(n, edges, -1);

        int m = edges.size();

        vector<int> critical, pcritical;

        for(int i=0; i<m; i++) {
            int temp = f(n, edges, i);
            int temp2 = f2(n, edges, i);
            if(temp>mini) critical.push_back(i);
            else if(temp2==mini) pcritical.push_back(i);
        }

        return {critical, pcritical};
    }
};
127284