<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: int, edges: List[List[int]]) -> List[List[int]]:
        mstWeight = 0
        dsu = DSU(n)
        modEdges = [[e[2], e[0], e[1], i] for i, e in enumerate(edges)]
        modEdges.sort()
        critical = []
        pseudo = []

        for cost, u, v, _ in modEdges:
            if dsu.union(u, v):
                mstWeight += cost
        
        for idx, (cost, u, v, edgeId) in enumerate(modEdges):
            # exclude weight
            excludeDSU = DSU(n)
            excludeWeight = 0
            for newIdx, (newCost, newU, newV, _) in enumerate(modEdges):
                if idx != newIdx and excludeDSU.union(newU, newV):
                    excludeWeight += newCost

            if excludeWeight > mstWeight or excludeDSU.maxSize < n:
                critical.append(edgeId)
            else:   # include weight, see if pseudocritical
                includeDSU = DSU(n)
                includeWeight = cost
                includeDSU.union(u, v)
                for newIdx, (newCost, newU, newV, _) in enumerate(modEdges):
                    if newIdx != idx and includeDSU.union(newU, newV):
                        includeWeight += newCost
                if includeWeight == mstWeight:
                    pseudo.append(edgeId)
            

        return [critical, pseudo]

class DSU:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.size = [1] * n
        self.maxSize = 1
    
    def find(self, u):
        if self.parent[u] != u:
            self.parent[u] = self.find(self.parent[u])
        return self.parent[u]

    def union(self, u, v):
        repU = self.find(u)
        repV = self.find(v)
        if repU == repV:
            return False
        if self.size[repU] < self.size[repV]:
            self.parent[repU] = repV
            self.size[repV] += self.size[repU]
            self.maxSize = max(self.maxSize, self.size[repV])
        else:
            self.parent[repV] = repU
            self.size[repU] += self.size[repV]
            self.maxSize = max(self.maxSize, self.size[repU])
        return True


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