<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]]:

        p = list(range(n))
        def getP(i, p):
            if p[i] == i:
                return i
            
            p[i] = p[getP(p[i], p)]
            return p[i]


        mst = 0
        ee = edges.copy()
        edges = sorted(list(enumerate(edges)),key = lambda x:x[1][-1])
        for i, e in edges:
            u, v, w = e
            if getP(u, p) != getP(v, p):
                mst += w
                p[getP(v, p)] = getP(u, p)
        
        
        def const(idx):
            mst_with = 0
            mst_wo = 0
            p_with = list(range(n))
            p_wo = list(range(n))

            ui, vi, wi = ee[idx]
            mst_with += wi
            p_with[getP(vi, p_with)] = getP(ui, p_with)
            count_with = 1
            count_wo = 0

            for i, e in edges:
                u, v, w = e
                if idx != i:
                    if getP(u, p_with) != getP(v, p_with):
                        mst_with += w
                        p_with[getP(v, p_with)] = getP(u, p_with)
                        count_with += 1
                    
                    if getP(u, p_wo) != getP(v, p_wo):
                        mst_wo += w
                        p_wo[getP(v, p_wo)] = getP(u, p_wo)
                        count_wo += 1
            
            if count_wo < n - 1:
                mst_wo = inf
            
            return mst_with, mst_wo
        
        psudo = []
        crit = []
        for i in range(len(edges)):
            wit, wo = const(i)
            # print(wit, wo, mst)
            if wit == wo == mst:
                psudo.append(i)
            if mst == wit < wo:
                crit.append(i)
        
        return [crit, psudo]

            
                
                


                

        
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