<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 DisjointSet{
    private:
        vector<int> parent, rank;
    public:
        DisjointSet(int& size){
            parent.resize(size);
            rank.resize(size, 0);

            for(int i = 0; i<size; ++i){
                parent[i] = i;
            }
        }

        int findParent(int u){
            //base case
            if(parent[u]==u) return u;
            return parent[u] = findParent(parent[u]);
        }

        bool UnionByRank(int u, int v){
            int ulp_u = findParent(u);
            int ulp_v = findParent(v);

            if(ulp_u==ulp_v) return false;
            if(rank[ulp_u] > rank[ulp_v]){
                parent[ulp_v] = ulp_u;
            }else if(rank[ulp_v] > rank[ulp_u]){
                parent[ulp_u] = ulp_v;
            }else{
                parent[ulp_v] = ulp_u;
                rank[ulp_u]++;
            }
            return true;
        }
};
static bool cmp (vector<int> &a, vector<int>& b){
    return a[2] < b[2];
}

class Solution {
public:
    int kruskal(int&n, int&m, vector<vector<int>>& edges, int blockedEdge, int takenEdge){
        int weight = 0;

        DisjointSet* ds = new DisjointSet(n);

        if(takenEdge!=-1){
            for(auto it: edges){
                if(it[3]==takenEdge){
                    weight += it[2];
                    ds->UnionByRank(it[0], it[1]);
                }
            }
        }

        for(auto it: edges){
            int u = it[0];
            int v = it[1];
            int w = it[2];
            int idx = it[3];

            if(idx==blockedEdge) continue;

            if(ds->UnionByRank(u, v)){
                weight += w;
            }else{
                continue;
            }
        }

        for(int i = 0; i<n; ++i){ //all nodes should be a part of mst
            if(ds->findParent(i)!=ds->findParent(0)) return INT_MAX;
        }

        return weight;
    }
    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(), cmp);

        int mst = kruskal(n, m, edges, -1, -1);

        vector<int> critical, pseudo;

        for(auto it: edges){
            if(kruskal(n, m, edges, it[3], -1) > mst){
                critical.push_back(it[3]);
            }else if(kruskal(n, m, edges, -1, it[3]) == mst){
                pseudo.push_back(it[3]);
            }
        }

        return {critical, pseudo};
    }
};
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