<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>

"""
Implementation of union find data structure to find MST
"""
class UnionFind:
    def __init__(self, n):
        self.parents = [i for i in range(n)]
        self.ranks = [1 for _ in range(n)]
        self.components = n

    """
    Find the root node of the components which contains given element (specified by 'val')
    """
    def find(self, val):
        if self.parents[val] != val:
            self.parents[val] = self.find(self.parents[val])
            
        return self.parents[val]

    """
    Union of two components if given elements belong to two different components
    Also return true of given elements belong to two different components else False
    """
    def union(self, val1, val2):
        parent1 = self.find(val1)
        parent2 = self.find(val2)
        if parent1 != parent2:
            if self.ranks[parent1] < self.ranks[parent1]:
                self.parents[parent1] = parent2
            else:
                self.parents[parent2] = parent1
                if self.ranks[parent2] == self.ranks[parent1]:
                    self.ranks[parent1] += 1
            self.components -= 1
            return True 

        return False 

    """
    Get the total number of components
    """
    def getComponents(self):
        return self.components


class Solution:
    """
    Create a MST where blockedEdge is skipped and preEdge will be always part of MST
    If blockedEdge is -1, nothing to skip
    If preEdge is -1, then no preEdge needs to be added in MST
    """
    def __getMST(self, edges, n, blockedEdge, preEdge = -1):
        unionFind = UnionFind(n)
        mstWeight = 0
        if preEdge != -1:
            mstWeight += edges[preEdge][2]
            unionFind.union(edges[preEdge][0], edges[preEdge][1])
        
        for i, edge in enumerate(edges):
            if i != blockedEdge and unionFind.union(edges[i][0], edges[i][1]):
                mstWeight += edges[i][2]

        return mstWeight if unionFind.getComponents() == 1 else float('inf')
        
    """
    Find Critical and pseudo-critical edges
    """
    def findCriticalAndPseudoCriticalEdges(self, n, edges):
        # Add position of edge in edges at the end of edge
        for i, edge in enumerate(edges):
            edge.append(i)

        edges.sort(key = lambda edge: edge[2])
        mstWeight = self.__getMST(edges, n, -1)

        # Find out critial and non-critical(pseudo critical) edges
        critical = []
        pseudoCritical = []
        for i in range(len(edges)):
            if mstWeight < self.__getMST(edges, n, i):
                critical.append(edges[i][3])
            elif mstWeight == self.__getMST(edges, n, -1, i):
                pseudoCritical.append(edges[i][3]) 
        
        return (critical, pseudoCritical)
820