<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]]:
        
        def make(node, parents, ranks):
            parents[node] = node
            ranks[node] = 0
    
        def find(node, parents, ranks):
            if parents[node] != node:
                parents[node] = find(parents[node], parents, ranks)
            return parents[node]
        
        def union(u, v, parents, ranks):
            u_rep = find(u, parents, ranks)
            v_rep = find(v, parents, ranks)

            if u_rep==v_rep:
                return False
            if ranks[u_rep] > ranks[v_rep]:
                parents[v_rep] = u_rep
            elif ranks[u_rep] < ranks[v_rep]:
                parents[u_rep] = v_rep
            else:
                parents[u_rep] = v_rep
                ranks[u_rep] += 1
            return True

        def kruskals(edges, nodes, initial_edge = None, ignored_edge = None):
            n = len(nodes)
            parents = dict()
            ranks = dict()
            for node in nodes:
                make(node,parents,ranks)

            total_weight = 0
            if initial_edge:
                node1, node2, weight = initial_edge
                union(node1, node2, parents, ranks)
                total_weight += weight
            
            for edge in edges:
                if edge != ignored_edge:
                    node1, node2, weight = edge
                    merged = union(node1, node2, parents, ranks)
                    if merged:
                        total_weight += weight
            
            root = None
            spanning = True
            for node in nodes:
                cur_root = find(node, parents, ranks)
                if root is None:
                    root = cur_root
                elif cur_root != root:
                    spanning = False
                    break
            
            return total_weight, spanning
        """
        solution outline:
        3 logical buckets: critical, pseudocritical, useless
        use kruskals algorithm to construct an MST
        go through every edge
                start an MST from this edge and use kruskals
                if weight is not min:
                    edge is useless
                else:
                    delete edge from graph
                    run kruskals on modified graph
                    if weight is min:
                        edge is pseudocritical
                    else:
                        edge is critical
        """
        nodes = set()
        for node1,node2, weight in edges:
            nodes.add(node1)
            nodes.add(node2)
        sorted_edges = sorted(edges, key = lambda x: x[2])
        min_weight, spanning = kruskals(sorted_edges, nodes)
        assert spanning

        critical = []
        pseudo = []
        for ind, edge in enumerate(edges):
            weight, spanning = kruskals(sorted_edges, nodes, edge, None)
            assert weight >= min_weight
            assert spanning
            if weight == min_weight:
                second_weight, spanning = kruskals(sorted_edges, nodes, None, edge)
                if spanning and second_weight==min_weight:
                    pseudo.append(ind)
                else:
                    critical.append(ind)
        return [critical, pseudo]





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