<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 {
    class UnionFind {
        vector<int> parents;
    public:
        int count;
        UnionFind(int n) {
            parents.resize(n);
            for (int i = 0; i < n; i++) {
                parents[i] = i;
            }
            count = n;
        }

        int find(int x) {
            if (parents[x] == x) {
                return x;
            }

            return parents[x] = find(parents[x]);
        }
        
        void merge(int x, int y) {
            int rootX = find(x);
            int rootY = find(y);
            if (rootX == rootY) {
                return;
            }
            parents[rootY] = rootX;
            count--;
        }
    };
public:
    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        vector<int> indexes(edges.size());
        for (int i = 0; i < indexes.size(); i++) {
            indexes[i] = i;
        }
        sort(indexes.begin(), indexes.end(), [&](int lhs, int rhs) {
            return edges[lhs][2] < edges[rhs][2];
        });

        int minCost = costOfMST(n, edges, indexes);

        vector<vector<int>> ans(2);
        for (auto index: indexes) {
            if (minCost < costOfMST(n, edges, indexes, index)) {
                ans[0].push_back(index);
            } else if (minCost == costOfMST(n, edges, indexes, -1, index)) {
                ans[1].push_back(index);
            }
        }
        return ans;
    }

    int costOfMST(int n, vector<vector<int>> &edges, vector<int> &indexes, int removed = -1, int added = -1) {
        int cost = 0;
        UnionFind uf(n);

        if (added != -1) {
            cost += edges[added][2];
            uf.merge(edges[added][0], edges[added][1]);
        }

        for (auto &index: indexes) {
            if (index == removed) {
                continue;
            }
            auto edge = edges[index];
            if (uf.find(edge[0]) == uf.find(edge[1])) {
                continue;
            }

            cost += edge[2];
            uf.merge(edge[0], edge[1]);
        }

        return uf.count == 1 ? cost : INT_MAX;
    }
};
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