<p>There is an undirected tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You are given the integer <code>n</code> and a 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>

<p>You are also given a <strong>0-indexed</strong> integer array <code>values</code> of length <code>n</code>, where <code>values[i]</code> is the <strong>value</strong> associated with the <code>i<sup>th</sup></code> node, and an integer <code>k</code>.</p>

<p>A <strong>valid split</strong> of the tree is obtained by removing any set of edges, possibly empty, from the tree such that the resulting components all have values that are divisible by <code>k</code>, where the <strong>value of a connected component</strong> is the sum of the values of its nodes.</p>

<p>Return <em>the <strong>maximum number of components</strong> in any valid split</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/07/example12-cropped2svg.jpg" style="width: 1024px; height: 453px;" />
<pre>
<strong>Input:</strong> n = 5, edges = [[0,2],[1,2],[1,3],[2,4]], values = [1,8,1,4,4], k = 6
<strong>Output:</strong> 2
<strong>Explanation:</strong> We remove the edge connecting node 1 with 2. The resulting split is valid because:
- The value of the component containing nodes 1 and 3 is values[1] + values[3] = 12.
- The value of the component containing nodes 0, 2, and 4 is values[0] + values[2] + values[4] = 6.
It can be shown that no other valid split has more than 2 connected components.</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/07/example21svg-1.jpg" style="width: 999px; height: 338px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]], values = [3,0,6,1,5,2,1], k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> We remove the edge connecting node 0 with 2, and the edge connecting node 0 with 1. The resulting split is valid because:
- The value of the component containing node 0 is values[0] = 3.
- The value of the component containing nodes 2, 5, and 6 is values[2] + values[5] + values[6] = 9.
- The value of the component containing nodes 1, 3, and 4 is values[1] + values[3] + values[4] = 6.
It can be shown that no other valid split has more than 3 connected components.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>values.length == n</code></li>
	<li><code>0 &lt;= values[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
	<li>Sum of <code>values</code> is divisible by <code>k</code>.</li>
	<li>The input is generated such that <code>edges</code> represents a valid tree.</li>
</ul>

from collections import defaultdict
from typing import List

class Solution:
    def maxKDivisibleComponents(self, n: int, edges: List[List[int]], values: List[int], k: int) -> int:
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)

        def dfs(node, parent):
            total = values[node]
            for neighbor in graph[node]:
                if neighbor != parent:
                    total += dfs(neighbor, node)
            if total % k == 0:
                self.components += 1
                return 0
            return total

        self.components = 0
        dfs(0, -1)
        return self.components
1205