<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] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with weight <code>w<sub>i</sub></code> in the tree.</p>

<p>You are also given a 2D integer array <code>queries</code> of length <code>m</code>, where <code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>. For each query, find the <strong>minimum number of operations</strong> required to make the weight of every edge on the path from <code>a<sub>i</sub></code> to <code>b<sub>i</sub></code> equal. In one operation, you can choose any edge of the tree and change its weight to any value.</p>

<p><strong>Note</strong> that:</p>

<ul>
	<li>Queries are <strong>independent</strong> of each other, meaning that the tree returns to its <strong>initial state</strong> on each new query.</li>
	<li>The path from <code>a<sub>i</sub></code> to <code>b<sub>i</sub></code> is a sequence of <strong>distinct</strong> nodes starting with node <code>a<sub>i</sub></code> and ending with node <code>b<sub>i</sub></code> such that every two adjacent nodes in the sequence share an edge in the tree.</li>
</ul>

<p>Return <em>an array </em><code>answer</code><em> of length </em><code>m</code><em> where</em> <code>answer[i]</code> <em>is the answer to the</em> <code>i<sup>th</sup></code> <em>query.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/11/graph-6-1.png" style="width: 339px; height: 344px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[0,1,1],[1,2,1],[2,3,1],[3,4,2],[4,5,2],[5,6,2]], queries = [[0,3],[3,6],[2,6],[0,6]]
<strong>Output:</strong> [0,0,1,3]
<strong>Explanation:</strong> In the first query, all the edges in the path from 0 to 3 have a weight of 1. Hence, the answer is 0.
In the second query, all the edges in the path from 3 to 6 have a weight of 2. Hence, the answer is 0.
In the third query, we change the weight of edge [2,3] to 2. After this operation, all the edges in the path from 2 to 6 have a weight of 2. Hence, the answer is 1.
In the fourth query, we change the weights of edges [0,1], [1,2] and [2,3] to 2. After these operations, all the edges in the path from 0 to 6 have a weight of 2. Hence, the answer is 3.
For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from a<sub>i</sub> to b<sub>i</sub>.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/11/graph-9-1.png" style="width: 472px; height: 370px;" />
<pre>
<strong>Input:</strong> n = 8, edges = [[1,2,6],[1,3,4],[2,4,6],[2,5,3],[3,6,6],[3,0,8],[7,0,2]], queries = [[4,6],[0,4],[6,5],[7,4]]
<strong>Output:</strong> [1,2,2,3]
<strong>Explanation:</strong> In the first query, we change the weight of edge [1,3] to 6. After this operation, all the edges in the path from 4 to 6 have a weight of 6. Hence, the answer is 1.
In the second query, we change the weight of edges [0,3] and [3,1] to 6. After these operations, all the edges in the path from 0 to 4 have a weight of 6. Hence, the answer is 2.
In the third query, we change the weight of edges [1,3] and [5,2] to 6. After these operations, all the edges in the path from 6 to 5 have a weight of 6. Hence, the answer is 2.
In the fourth query, we change the weights of edges [0,7], [0,3] and [1,3] to 6. After these operations, all the edges in the path from 7 to 4 have a weight of 6. Hence, the answer is 3.
For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from a<sub>i</sub> to b<sub>i</sub>.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt; n</code></li>
	<li><code>1 &lt;= w<sub>i</sub> &lt;= 26</code></li>
	<li>The input is generated such that <code>edges</code> represents a valid tree.</li>
	<li><code>1 &lt;= queries.length == m &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
</ul>

class WeightEquilibrium:
    def __init__(self, n):
        self.m = 1 + math.ceil(math.log2(2 * 1e4))
        self.n = n
        self.total_w = 27
        self.root_node = 0

        self.weights = [None for i in range(n)]
        self.parents = [[-1 for i in range(n)] for j in range(self.m)]
        self.depth = [0 for i in range(n)]

        self.graph = collections.defaultdict(list)

    def create_graph(self, edges):
        for a, b, w in edges:
            self.graph[a].append([b, w])
            self.graph[b].append([a, w])

    def initialize_graph_params(self):
        self.weights[self.root_node] = [0] * self.total_w
        self._dfs(self.root_node, -1, 0)

    def _dfs(self, curr, prev, depth):
        self.parents[0][curr] = prev
        self.depth[curr] = depth

        for nei, wt in self.graph[curr]:
            if nei == prev:
                continue
            self.weights[nei] = self.weights[curr].copy()
            self.weights[nei][wt] += 1
            self._dfs(nei, curr, depth + 1)

    def build_parents(self):
        for i in range(1, self.m):
            for j in range(self.n):
                if self.parents[i - 1][j] == -1:
                    self.parents[i][j] = -1
                else:
                    self.parents[i][j] = self.parents[i - 1][self.parents[i - 1][j]]

    def find_lca(self, u, v):
        if self.depth[v] > self.depth[u]:
            u, v = v, u
        diff = self.depth[u] - self.depth[v]
        for i in range(self.m - 1, -1, -1):
            if diff & (1 << i):
                u = self.parents[i][u]

        if u == v:
            return u

        for i in range(self.m - 1, -1, -1):
            if self.parents[i][u] != self.parents[i][v]:
                u, v = self.parents[i][u], self.parents[i][v]
        return self.parents[0][u]

    def edges_between(self, u, v, lca):
        return self.depth[u] + self.depth[v] - 2 * self.depth[lca]

    def weights_between(self, u, v, lca, w):
        return self.weights[u][w] + self.weights[v][w] - 2 * self.weights[lca][w]

    def process_query(self, query):
        node1, node2 = query
        lca_node = self.find_lca(node1, node2)

        no_of_edges = self.edges_between(node1, node2, lca_node)
        min_operations = no_of_edges

        for w in range(1, self.total_w):
            weights_w = self.weights_between(node1, node2, lca_node, w)
            min_operations = min(min_operations, no_of_edges - weights_w)

        return min_operations

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

        wt_eq = WeightEquilibrium(n)
        wt_eq.create_graph(edges)

        wt_eq.initialize_graph_params()
        wt_eq.build_parents()

        res = []
        for query in queries:
            res.append(wt_eq.process_query(query))

        return res
30100