<p>There is an undirected weighted graph with <code>n</code> vertices labeled from <code>0</code> to <code>n - 1</code>.</p>

<p>You are given the integer <code>n</code> and an array <code>edges</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between vertices <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with a weight of <code>w<sub>i</sub></code>.</p>

<p>A walk on a graph is a sequence of vertices and edges. The walk starts and ends with a vertex, and each edge connects the vertex that comes before it and the vertex that comes after it. It&#39;s important to note that a walk may visit the same edge or vertex more than once.</p>

<p>The <strong>cost</strong> of a walk starting at node <code>u</code> and ending at node <code>v</code> is defined as the bitwise <code>AND</code> of the weights of the edges traversed during the walk. In other words, if the sequence of edge weights encountered during the walk is <code>w<sub>0</sub>, w<sub>1</sub>, w<sub>2</sub>, ..., w<sub>k</sub></code>, then the cost is calculated as <code>w<sub>0</sub> &amp; w<sub>1</sub> &amp; w<sub>2</sub> &amp; ... &amp; w<sub>k</sub></code>, where <code>&amp;</code> denotes the bitwise <code>AND</code> operator.</p>

<p>You are also given a 2D array <code>query</code>, where <code>query[i] = [s<sub>i</sub>, t<sub>i</sub>]</code>. For each query, you need to find the minimum cost of the walk starting at vertex <code>s<sub>i</sub></code> and ending at vertex <code>t<sub>i</sub></code>. If there exists no such walk, the answer is <code>-1</code>.</p>

<p>Return <em>the array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> denotes the <strong>minimum</strong> cost of a walk for query </em><code>i</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,-1]</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/31/q4_example1-1.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 351px; height: 141px;" />
<p>To achieve the cost of 1 in the first query, we need to move on the following edges: <code>0-&gt;1</code> (weight 7), <code>1-&gt;2</code> (weight 1), <code>2-&gt;1</code> (weight 1), <code>1-&gt;3</code> (weight 7).</p>

<p>In the second query, there is no walk between nodes 3 and 4, so the answer is -1.</p>

<p><strong class="example">Example 2:</strong></p>
</div>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[0]</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/01/31/q4_example2e.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 211px; height: 181px;" />
<p>To achieve the cost of 0 in the first query, we need to move on the following edges: <code>1-&gt;2</code> (weight 1), <code>2-&gt;1</code> (weight 6), <code>1-&gt;2</code> (weight 1).</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>0 &lt;= w<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= query.length &lt;= 10<sup>5</sup></code></li>
	<li><code>query[i].length == 2</code></li>
	<li><code>0 &lt;= s<sub>i</sub>, t<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>s<sub>i</sub> !=&nbsp;t<sub>i</sub></code></li>
</ul>

from functools import reduce
import operator

class Solution:
    def minimumCost(self, n: int, edges: List[List[int]], query: List[List[int]]) -> List[int]:
        
        graph = {}
        
        for u,v,w in edges:
            if u not in graph:
                graph[u] = []
            if v not in graph:
                graph[v] = []
            graph[u].append([v,w])
            graph[v].append([u,w])
        
        visited = set()
        
        def dfs(node,graph):
            res = set()
            nodes_in_graph = {node}
            stack = [node]
            
            while stack:
                v = stack.pop()
                if v not in visited:
                    visited.add(v)
                    for neighbor, weight in graph[v]:
                        if neighbor not in visited:
                            stack.append(neighbor)
                            nodes_in_graph.add(neighbor)
                            if (v,neighbor,weight) not in res and (neighbor,v,weight) not in res:
                                res.add((v,neighbor,weight))
            return res,nodes_in_graph
                                
        
        connected_graphs = []
        nodes_in_graphs = []
        
        for node in graph:
            if node not in visited:
                component,nodes_in_graph_component = dfs(node,graph)
                connected_graphs.append(component)
                nodes_in_graphs.append(nodes_in_graph_component)
                
        def find_minimum_weight(edges):
            connected_set = set()
            
            for vertex_1,vertex_2,weight in edges:
                connected_set.add(vertex_1)
                connected_set.add(vertex_2)
            
            return reduce(operator.and_, [edge[-1] for edge in edges])
        
        min_list = []
        
        for edges in connected_graphs:
            min_list.append(find_minimum_weight(edges))
        
        ans = []
        added = False
        
        for u, v in query:
            
            for i in range(len(nodes_in_graphs)):
                if u in  nodes_in_graphs[i]and v in nodes_in_graphs[i]:
                    ans.append(min_list[i])
                    added = True
                    
            if not added:
                ans.append(-1)
                
            added = False
                    
        
        return ans
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