<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 {
        int n;
        vector<int> parent, rank;
    public:
        dsu(int n) {
            this->n = n;
            parent.resize(n);
            rank.resize(n, 1);
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }
        int find(int a) {
            if (a == parent[a]) return a;
            return parent[a] = find(parent[a]);
        }
        bool unionr(int a, int b) {
            a = find(a);
            b = find(b);
            if (a == b) return false;
            if (rank[a] > rank[b]) parent[b] = a;
            else parent[a] = b;
            if (rank[a] == rank[b]) rank[a]++;
            return true;
        }
        void reset() {
            for (int i = 0; i < n; i++) {
                rank[i] = 1;
                parent[i] = i;
            }
        }
    };

private:
    int N;
    dsu *ds;
    int edgecount;
    vector<vector<int>> myedges;

    int findmst(int avoidme, bool alreadypicked) {
        int ans = 0;
        int totalop = 0;
        if (alreadypicked) totalop++;
        for (int i = 0; i < edgecount; i++) {
            if (totalop == N - 1) break;
            if (myedges[i][3] == avoidme) continue;
            if (ds->unionr(myedges[i][1], myedges[i][2])) {
                ans += myedges[i][0];
                totalop++;
            }
        }
        ds->reset();
        return totalop == N - 1 ? ans : INT_MAX;
    }

public:
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        N = n;
        edgecount = edges.size();
        myedges.clear();
        myedges.assign(edgecount, {});

        for (int i = 0; i < edgecount; i++) {
            myedges[i].push_back(edges[i][2]);
            myedges[i].push_back(edges[i][0]);
            myedges[i].push_back(edges[i][1]);
            myedges[i].push_back(i);
        }
        
        sort(myedges.begin(), myedges.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[0] < b[0];
        });

        ds = new dsu(N);
        int mst = findmst(-1, false);
        if (mst == INT_MAX) {
            delete ds;
            return {};
        }

        vector<int> critical, pcritical;
        for (int i = 0; i < edgecount; i++) {
            int cmst = findmst(i, false);
            if (cmst > mst) {
                critical.push_back(i);
            } else if (cmst == mst) {
                ds->unionr(edges[i][0], edges[i][1]);
                cmst = edges[i][2] + findmst(-1, true);
                if (cmst == mst) {
                    pcritical.push_back(i);
                }
            }
            ds->reset();
        }

        delete ds;
        return {critical, pcritical};
    }
};

15853