<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{
    vector<int> parent, size;
    public:

    DisjointSet(int v){
        parent.resize(v+1);
        size.resize(v+1);

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

        for(int i = 0 ; i <= v ; i++){
            size[i] = 1;
        }
    }

    int findParent(int node){
        if(parent[node] == node){
            return node;
        }

        return parent[node] = findParent(parent[node]);
    }

    void unionBySize(int u, int v){
        int up = findParent(u);
        int vp = findParent(v);

        if(up == vp){
            return;
        }

        if(size[up] < size[vp]){
            size[vp] += size[up];
            parent[up] = vp;
        }
        else{
            size[up] += size[vp];
            parent[vp] = up;
        }
    }
};

class Solution {
    int getMstWeight(int n, map<int, vector<pair<int, int>>> &maps, pair<int, int> remove, pair<int, pair<int, int>> add){
        DisjointSet ds(n);
        int cost = 0;

        if(add.first != -1){
            ds.unionBySize(add.first, add.second.first);
            cost = add.second.second;
        }
        for(auto it: maps){
            for(auto num: it.second){
                int u = num.first;
                int v = num.second;

                if(remove.first == u && remove.second == v) continue; 


                int up = ds.findParent(u);
                int vp = ds.findParent(v);


                if(up == vp) continue;

                ds.unionBySize(u, v);
                cost += it.first;
            }
        }

        int cnt = 0;
        for(int i = 0 ; i < n ; i++){
            if(ds.findParent(i) == i){
                cnt++;
            }
        }

        if(cnt > 1) {
            return INT_MAX;
        }

        return cost;
    }
public:
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        map<int, vector<pair<int, int>>> maps;

        for(auto it: edges){
            maps[it[2]].push_back({it[0], it[1]});
        }

        int m = edges.size();
        int mstWeight = getMstWeight(n, maps, {-1, -1}, {-1, {-1, 0}});
        
        vector<vector<int>> ans(2);
        for(int i = 0 ; i < m ; i++){
            int u = edges[i][0];
            int v = edges[i][1];
            int w = edges[i][2];

            int curWeight = getMstWeight(n, maps, {u, v}, {-1, {-1, 0}});
            if(curWeight > mstWeight){
                ans[0].push_back(i);
            }
            else if(getMstWeight(n, maps, {-1, -1}, {u, {v, w}}) == mstWeight){
                ans[1].push_back(i);
            }
        }

        return ans;
    }
};
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