<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:
    class Disjoint {
    public:
        vector<int> parent, size;
        int comp, maxSize = 0;
        Disjoint(int n) {
            comp = n;
            parent.resize(n);
            size.resize(n);
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }

        int findPar(int x) {
            if (x == parent[x])
                return x;
            return findPar(parent[x]);
        }

        void unite(int a, int b) {
            int parA = findPar(a), parB = findPar(b);
            if (parA == parB)
                return;
            if (size[parA] > size[parB]) {
                parent[parB] = parA;
                size[parA] += size[parB];
                maxSize = max(size[parA], maxSize);
            } else {
                parent[parA] = parB;
                size[parB] += size[parA];
                maxSize = max(size[parB], maxSize);
            }
        }
    };

    vector<vector<int>>
    findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>> edges) {
        int mstCost = 0, eSize = edges.size();
        vector<vector<int>> res(2);
        Disjoint dsM(n);
        for (int i = 0; i < eSize; i++)
            edges[i].push_back(i);
        sort(edges.begin(), edges.end(),
             [](auto& a, auto& b) { return a[2] < b[2]; });

        for (int i = 0; i < eSize; i++) {
            auto e = edges[i];
            if (dsM.findPar(e[0]) != dsM.findPar(e[1])) {
                mstCost += e[2];
                dsM.unite(e[0], e[1]);
            }
        }

        for (auto ed : edges) {
            // trying without curr edge
            Disjoint ds(n);
            int newCst = 0;
            for (int i = 0; i < eSize; i++) {
                auto e = edges[i];
                if (ed[3] != e[3] && ds.findPar(e[0]) != ds.findPar(e[1])) {
                    newCst += e[2];
                    ds.unite(e[0], e[1]);
                }
            }
            // if all the nodes are connected and new cost is more than the mst
            // then it is critical node
            if (newCst > mstCost || ds.maxSize != n) {
                res[0].push_back(ed[3]);
                continue;
            }

            // trying with forcing the edge
            Disjoint ds1(n);
            ds1.unite(ed[0], ed[1]);
            newCst = ed[2];
            for (int i = 0; i < eSize; i++) {
                auto e = edges[i];
                if (ed[3] != e[3] && ds1.findPar(e[0]) != ds1.findPar(e[1])) {
                    newCst += e[2];
                    ds1.unite(e[0], e[1]);
                }
            }
            // if the new cost is same as the mst then its a pseudo edge
            if (newCst == mstCost)
                res[1].push_back(ed[3]);
        }
        return res;
    }
};
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