<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>

import heapq

class Solution:
    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:

        graph = {i: [] for i in range(n)}

        for u, v, w in edges:
            graph[u].append((v, w))
            graph[v].append((u, w))

        def prim_mst(graph, dist, visited):
            pq = [(0, 0)]

            while pq:
                edge_w, node = heapq.heappop(pq)

                if node not in visited:

                    for neighbor, w in graph[node]:

                        if (w < dist[neighbor] or dist[neighbor] == -1) and neighbor not in visited:
                            dist[neighbor] = w
                            heapq.heappush(pq, (dist[neighbor], neighbor))

                    visited.add(node)

            return sum(dist[i] for i in range(n))

        dist = {i : -1 for i in range(n)}
        dist[0] = 0
        visited = set()
        mst_weight = prim_mst(graph, dist, visited)

        critical = []
        pseudo = []

        for i in range(len(edges)):
            g = {j: [] for j in range(n)}

            for j, (u, v, w) in enumerate(edges):
                if i == j:
                    continue
                g[u].append((v, w))
                g[v].append((u, w))

            dist = {i : -1 for i in range(n)}
            dist[0] = 0
            visited = set()

            temp_mst = prim_mst(g, dist, visited)

            if temp_mst != mst_weight:
                critical.append(i)
            else:
                edge = edges[i]

                g = {j: [] for j in range(n)}

                for j, (u, v, w) in enumerate(edges):
                    if i == j:
                        g[u].append((v, 0))
                        g[v].append((u, 0))
                    else:
                        g[u].append((v, w))

                        g[v].append((u, w))
                
                
                dist = {i : -1 for i in range(n)}
                dist[0] = 0
                visited = set()

                temp_mst = prim_mst(g, dist, visited) + edge[2]


                if temp_mst == mst_weight:
                    pseudo.append(i)
                

                

        return [critical, pseudo]





            
        
        
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