<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:
    int find(vector<int> &uf, int node) {
        if (node == uf[node])
            return uf[node];
        uf[node] = find(uf, uf[node]);
        return uf[node];
    }
    int BuildMST(vector<int> &uf, vector<vector<int>> &edges, set<int> &mst) {
        int res = 0;
        for (int i = 0 ; i < edges.size() ; i++) {
            if (edges[i][4])
                continue;
            if (find(uf, edges[i][0]) == find(uf, edges[i][1]))
                continue;
            uf[find(uf, edges[i][0])] = uf[find(uf, edges[i][1])];
            res += edges[i][2];
            mst.insert(i);
        }
        if (mst.size() + 1 < uf.size())
            return -1;
        return res;
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<int> uf(n);
        for (int i = 0 ; i < n ; i++)
            uf[i] = i;
        for (int i = 0 ; i < edges.size() ; i++) {
            edges[i].push_back(i);
            edges[i].push_back(0);
        }
        sort(edges.begin(), edges.end(), [](vector<int> &l, vector<int> &r){return l[2] < r[2];});
        set<int> mst;
        int m = BuildMST(uf, edges, mst);
        vector<vector<int>> res(2, vector<int>());
        for (int i : mst) {
            for (int j = 0 ; j < n ; j++) {
                uf[j] = j;
            }
            edges[i][4] = 1;
            set<int> tmp;
            int t = BuildMST(uf, edges, tmp);
            if (t != m) {
                res[0].push_back(edges[i][3]);
            } else {
                res[1].push_back(edges[i][3]);
            }
            edges[i][4] = 0;
        }
        for (int i = 0 ; i < edges.size() ; i++) {
            if (mst.contains(i))
                continue;
            for (int j = 0 ; j < n ; j++) {
                uf[j] = j;
            }
            uf[edges[i][0]] = uf[edges[i][1]];
            set<int> tmp;
            tmp.insert(i);
            int t = BuildMST(uf, edges, tmp);
            if (edges[i][2] + t == m) {
                res[1].push_back(edges[i][3]);
            }
        }
        return res;
    }
};
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