<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 findparent(int x, vector<int> &parent) {
        if(parent[x] == -1) {
            return x;
        }
        return parent[x] = findparent(parent[x], parent);
    }

    void unionset(int x, int y, vector<int> &parent, vector<int>&rank) {
        int s1 = findparent(x, parent);
        int s2 = findparent(y, parent);
        if(s1 != s2) {
            if(rank[s1] < rank[s2]) {
                parent[s1] = s2;
                rank[s2] += rank[s1];
            } else {
                parent[s2] = s1;
                rank[s1] += rank[s2];
            }
        }
    }

    static bool compare(vector<int>&u, vector<int>&v) {
        return u[2] < v[2];
    }

    int mst(int n, vector<vector<int>>& edges) {
        vector<int> parent(n, -1);
        vector<int> rank(n, 1);
        int ans = 0;
        int count = 0;
        for(auto& edge : edges) {
            int x = edge[0];
            int y = edge[1];
            int w = edge[2];
            int s1 = findparent(x, parent);
            int s2 = findparent(y, parent);
            if(s1 != s2) {
                ans += w;
                unionset(x, y, parent, rank);
                count++;
            }
        }
        return (count == n-1) ? ans : INT_MAX;
    }

    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        int m = edges.size();
        for(int i = 0; i < m; i++) {
            edges[i].push_back(i);
        }
        sort(edges.begin(), edges.end(), compare);
        vector<int> critical;
        vector<int> pseudocritical;
        int ans = mst(n, edges);

        for(int i = 0; i < m; i++) {
            vector<vector<int>> temp(edges);
            temp.erase(temp.begin() + i);
            int new_ans = mst(n, temp);
            
            if(new_ans > ans) {
                critical.push_back(edges[i][3]);
            } else {
                int curr_ans = 0;
                vector<int> parent(n, -1);
                vector<int> rank(n, 1);
                unionset(edges[i][0], edges[i][1], parent, rank);
                curr_ans = edges[i][2];

                for(auto& edge : edges) {
                    if(edge != edges[i]) {
                        int x = edge[0];
                        int y = edge[1];
                        int w = edge[2];
                        int s1 = findparent(x, parent);
                        int s2 = findparent(y, parent);
                        if(s1 != s2) {
                            curr_ans += w;
                            unionset(x, y, parent, rank);
                        }
                    }
                }

                if(curr_ans == ans) {
                    pseudocritical.push_back(edges[i][3]);
                }
            }
        }

        return {critical, pseudocritical};
    }
};
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