<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>

from typing import List, Dict, Tuple
from collections import defaultdict, deque, Counter

class Solution:
    def __init__(self):
        self.parent = []
        self.depth = []
        self.weights = []
        self.edges_info = defaultdict(list)
        
    def bfs(self, n: int):
        self.parent = [-1] * n
        self.depth = [0] * n
        self.weights = [{} for _ in range(n)]
        queue = deque([0])
        visited = [False] * n
        visited[0] = True
        
        while queue:
            node = queue.popleft()
            for neighbor, w in self.edges_info[node]:
                if not visited[neighbor]:
                    visited[neighbor] = True
                    self.parent[neighbor] = node
                    self.depth[neighbor] = self.depth[node] + 1
                    self.weights[neighbor] = self.weights[node].copy()
                    self.weights[neighbor][w] = self.weights[neighbor].get(w, 0) + 1
                    queue.append(neighbor)

    def lca(self, u: int, v: int) -> int:
        if self.depth[u] < self.depth[v]:
            u, v = v, u
        while self.depth[u] > self.depth[v]:
            u = self.parent[u]
        while u != v:
            u = self.parent[u]
            v = self.parent[v]
        return u

    def minOperationsQueries(self, n: int, edges: List[List[int]], queries: List[List[int]]) -> List[int]:
        # Fill edge info
        for u, v, w in edges:
            self.edges_info[u].append((v, w))
            self.edges_info[v].append((u, w))
        
        # Process input using BFS to calculate depths, parents and prefix weight sums
        self.bfs(n)
        
        # Process each query
        results = []
        for a, b in queries:
            lca = self.lca(a, b)
            total_steps = self.depth[a] + self.depth[b] - 2 * self.depth[lca]
            
            # Use prefix sums on edge counts to determine the frequency of each weight on the path
            weight_count = Counter()
            for node in [a, b]:
                for weight in self.weights[node]:
                    weight_count[weight] += self.weights[node][weight]
            
            # Reverse the counts for LCA to subtract it twice
            for weight in self.weights[lca]:
                if weight in weight_count:
                    weight_count[weight] -= 2 * self.weights[lca][weight]

            # The number of operations required is finding the max repetitions of any weight and subtracting that from total steps
            max_weight_freq = max(weight_count.values(), default=0)
            min_operations = total_steps - max_weight_freq
            results.append(min_operations)

        return results
36900