<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 {

public:
    vector<int> parent;
    vector<int> Size;

    DisjointSet(int n) {
        parent.resize(n, 0);
        Size.resize(n, 1);
        for (int i = 0; i < n; i++)
            parent[i] = i;
    }

    void UnionBySize(int u, int v) {
        int pr_u = findUpr(u);
        int pr_v = findUpr(v);
        if (pr_u == pr_v)
            return;
        if (Size[pr_u] > Size[pr_v]) {
            parent[pr_v] = pr_u;
            Size[pr_u] += Size[pr_v];
        } else {
            parent[pr_u] = pr_v;
            Size[pr_v] += Size[pr_u];
        }
    }

    int findUpr(int u) {
        if (parent[u] == u)
            return u;
        return parent[u] = findUpr(parent[u]);
    }
};
class Solution {
public:
    int solve(int i, auto& edges, int n) {

        int mn = 0;
        DisjointSet ds1(n);
        int k = n;

        for (auto it : edges) {
            int wt = it[0];
            int u = it[1];
            int v = it[2];
            int j = it[3];
            if (i == j) {
                mn += wt;
                ds1.UnionBySize(u, v);
                k--;
            }
        }
        for (auto it : edges) {
            int wt = it[0];
            int u = it[1];
            int v = it[2];
            int j = it[3];

            if ((i != j) && (ds1.findUpr(u) != ds1.findUpr(v))) {
                k--;
                ds1.UnionBySize(u, v);
                // cout<<i<<" "<<j<<" "<<u<<" "<<v<<endl;
                mn += wt;
            }
        }
        // cout<<i<<" "<<mn<<endl;
        if (k == 1)
            return mn;
        return 1e9;
    }
    int solve2(int i, auto& edges, int n) {

        int mn = 0;
        DisjointSet ds1(n);
        int k = n;

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

            if ((i != j) && (ds1.findUpr(u) != ds1.findUpr(v))) {
                k--;
                ds1.UnionBySize(u, v);
                // cout<<i<<" "<<j<<" "<<u<<" "<<v<<endl;
                mn += wt;
            }
        }
        // cout<<i<<" "<<mn<<endl;
        if (k == 1)
            return mn;
        return 1e9;
    }
    vector<vector<int>>
    findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        int m = edges.size();
        for (int i = 0; i < edges.size(); i++) {
            int u = edges[i][0];
            int v = edges[i][1];
            int wt = edges[i][2];

            edges[i][0] = wt;
            edges[i][1] = u;
            edges[i][2] = v;
            edges[i].push_back(i);
        }
        sort(edges.begin(), edges.end());
        int mn = 0;
        DisjointSet ds(n);
        for (auto it : edges) {
            int wt = it[0];
            int u = it[1];
            int v = it[2];
            if (ds.findUpr(u) != ds.findUpr(v)) {
                ds.UnionBySize(u, v);
                mn += wt;
            }
            // cout<<wt<<" "<<u<<" "<<v<<" "<<it[3]<<endl;
        }
        // cout<<mn<<endl;
        vector<int> cr;
        vector<int> pcr;
        for (int i = 0; i < m; i++) {
            int temp = solve2(i, edges, n);
            if (temp > mn) {
                cr.push_back(i);
            } else {

                temp = solve(i, edges, n);

                if (temp == mn) {
                    pcr.push_back(i);
                }
            }
        }
       
        vector<vector<int>> ans;
        ans.push_back(cr);
        ans.push_back(pcr);
        return ans;
    }
};
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