<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 findpar(vector<int>& parent, int node) {
        if (node == parent[node]) {
            return node;
        }
        return parent[node] = findpar(parent, parent[node]);
    }
    void unionbysize(vector<int>& size, vector<int>& parent, int u, int v) {
        int ulp_u = findpar(parent, u);
        int ulp_v = findpar(parent, v);
        if (ulp_u == ulp_v) {
            return;
        }
        if (size[ulp_u] >= size[ulp_v]) {
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v];
        } else {
            parent[ulp_u] = ulp_v;
        }
    }
    static bool compare(vector<int>& a, vector<int>& b) { return a[2] < b[2]; }
    vector<vector<int>>
    findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<int> parent(n);
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
        vector<int> size(n, 1);
        unordered_map<int, int> m;
        for (int i = 0; i < edges.size(); i++) {
            edges[i].push_back(i);
        }
        sort(edges.begin(), edges.end(), compare);
        int ans = 0;
        for (int i = 0; i < edges.size(); i++) {
            if (findpar(parent, edges[i][0]) == findpar(parent, edges[i][1])) {
                continue;
            }
            ans += edges[i][2];
            unionbysize(size, parent, edges[i][0], edges[i][1]);
        }

        unordered_map<int,int>r;

        for (int i = 0; i < edges.size(); i++) {
            unordered_map<int, int> temp;
            int canans = 0;
            vector<int> parent1(n);
            vector<int> size1(n, 1);
            for (int i = 0; i < parent1.size(); i++) {
                parent1[i] = i;
            }
            for (int j = 0; j < edges.size(); j++) {
                if (i != j) {
                    if (findpar(parent1, edges[j][0]) ==
                        findpar(parent1, edges[j][1])) {
                        continue;
                    }
                    unionbysize(size1, parent1, edges[j][0], edges[j][1]);
                    canans += edges[j][2];
                    temp[edges[j][3]] = 1;
                }
            }

            if (canans != ans) {
                
                r[edges[i][3]]=1;

            } else {
                for (auto k : temp) {
                    
                    m[k.first] = 1;
                }
            }
        }
        vector<int>req;
        vector<int> notreq;
        
        
        for (int i = 0; i < edges.size(); i++) {

            if (m.find(edges[i][3]) == m.end() && r.find(edges[i][3])==r.end()) {
                vector<int> parent1(n);
                vector<int> size1(n, 1);
                for (int i = 0; i < parent1.size(); i++) {
                    parent1[i] = i;
                }
                int canans = 0;
                unionbysize(size1, parent1, edges[i][0], edges[i][1]);
                canans+=edges[i][2];

                for (int j = 0; j < edges.size(); j++) {
                    if (i != j) {
                        if (findpar(parent1, edges[j][0]) ==
                            findpar(parent1, edges[j][1])) {
                            continue;
                        }
                        unionbysize(size1, parent1, edges[j][0], edges[j][1]);
                        canans += edges[j][2];
                        
                    }
                }
                if(canans==ans){
                    notreq.push_back(edges[i][3]);

                }
            }
            if(r.find(edges[i][3])!=r.end()){
                req.push_back(edges[i][3]);
            }
            else if(m.find(edges[i][3])!=m.end()){
                notreq.push_back(edges[i][3]);
            }
        }
        vector<vector<int>> res;
        res.push_back(req);
        res.push_back(notreq);
        return res;
    }
};
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