<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:
    struct union_find {
        vector<int> parent;
        vector<int> size; 
        union_find(int n){
            parent=vector<int>(n);
            size=vector<int>(n, 1);
            for(int i=0;i<n;i++) parent[i]=i;
        }
        int _find(int x){
            if(parent[x]!=x) parent[x] = _find(parent[x]); 
            return parent[x];
        }
        int _union(int x, int y) {
            int px=_find(x),py = _find(y);
            if(px==py) return 0;
            if(size[px]>size[py]) swap(px,py);
            parent[px]=py;
            size[py]+=size[px];
            return 1;
        }

    };

    auto kruskal(struct union_find uf, vector<vector<int>>& edges, int remove){
        struct result {int cost, edges_cnt;};
        int cost=0,edges_cnt=0;
        for(int i=0;i<edges.size();i++){
            if(i==remove) continue;
            int x=edges[i][0];
            int y=edges[i][1];
            int w=edges[i][2];
            if(uf._union(x,y)){
                edges_cnt++;
                cost+=w;
            }
        }
        return result{cost,edges_cnt};
    }
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        for(int i=0;i<edges.size();i++) edges[i].push_back(i);
        sort(edges.begin(),edges.end(),[](auto a, auto b) {
            return a[2]<b[2];
        });
        
        union_find uf(n);
        auto[cost,edge_cnt]=kruskal(uf,edges,-1);

        vector<vector<int>> ret(2);
        for(int i=0;i<edges.size();i++) {
            union_find uf_new(n);
            auto res=kruskal(uf_new,edges,i);
            if(res.cost>cost || res.edges_cnt<n-1){
                ret[0].push_back(edges[i][3]);
                continue;
            } 
            uf_new=union_find(n);
            uf_new._union(edges[i][0],edges[i][1]);
            res=kruskal(uf_new,edges,-1);
            res.cost+=edges[i][2];
            res.edges_cnt++;
            if(res.cost==cost && res.edges_cnt==n-1){
                ret[1].push_back(edges[i][3]);
            }
        }
        return ret;
    }
};
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