<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:
    def findCriticalAndPseudoCriticalEdges(self, n, edges):
        def find(parent, i):
            if parent[i] != i:
                parent[i] = find(parent, parent[i])
            return parent[i]

        def union(parent, rank, x, y):
            root_x = find(parent, x)
            root_y = find(parent, y)
            if root_x != root_y:
                if rank[root_x] > rank[root_y]:
                    parent[root_y] = root_x
                elif rank[root_x] < rank[root_y]:
                    parent[root_x] = root_y
                else:
                    parent[root_y] = root_x
                    rank[root_x] += 1

        def kruskal(edges, skip, include):
            parent = list(range(n))
            rank = [0] * n
            total_weight = 0

            if include != -1:
                u, v, w, _ = edges[include]
                union(parent, rank, u, v)
                total_weight += w

            for i, edge in enumerate(edges):
                if i == skip:
                    continue
                u, v, w, _ = edge
                root_u = find(parent, u)
                root_v = find(parent, v)
                if root_u != root_v:
                    union(parent, rank, u, v)
                    total_weight += w

            for i in range(n):
                find(parent, i)
            return total_weight if len(set(parent)) == 1 else float('inf')

        edges = [edge + [i] for i, edge in enumerate(edges)]
        edges.sort(key=lambda x: x[2])

        mst_weight = kruskal(edges, -1, -1)
        critical, pseudo_critical = [], []

        for i, edge in enumerate(edges):
            u, v, w, _ = edge
            if kruskal(edges, i, -1) > mst_weight:
                critical.append(edge[3])
            elif kruskal(edges, -1, i) == mst_weight:
                pseudo_critical.append(edge[3])

        return [critical, pseudo_critical]

# Example usage:
solution = Solution()
n = 5
edges = [[0,1,1],[1,2,1],[2,3,2],[0,3,2],[0,4,3],[3,4,3],[1,4,6]]
print(solution.findCriticalAndPseudoCriticalEdges(n, edges))  # Output: [[0,1], [2,3,4,5]]
758