<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 Solution:
    def minOperationsQueries(self, n: int, edges: List[List[int]], queries: List[List[int]]) -> List[int]:
        tree = [[] for _ in range(n)]
        for u, v, w in edges: 
            tree[u].append((v, w))
            tree[v].append((u, w))
        m = int(log2(n)) + 2
        lift = [[-1]*m for _ in range(n)] # binary lifting 
        freq = [[0]*27 for _ in range(n)]
        depth = [-1]*n 
        stack = [(0, -1, 0)]
        while stack: 
            u, p, d = stack.pop()
            depth[u] = d 
            for v, w in tree[u]: 
                if v != p: 
                    lift[v][0] = u
                    freq[v][:] = freq[u][:]
                    freq[v][w] += 1
                    for j in range(1, m): 
                        if lift[v][j-1] == -1: break 
                        lift[v][j] = lift[lift[v][j-1]][j-1]
                    stack.append((v, u, d+1))
        
        def lca(u, v): 
            """Return lowest common ancestor via binary lifting."""
            if depth[u] > depth[v]: u, v = v, u
            for i in range(m): 
                if depth[v]-depth[u] & 1<<i: v = lift[v][i]
            if u == v: return u 
            for i in range(m-1, -1, -1): 
                if lift[u][i] != lift[v][i]: u, v = lift[u][i], lift[v][i]
            return lift[u][0]
        
        ans = []
        for u, v in queries: 
            k = lca(u, v)
            count = [freq[u][w] + freq[v][w] - 2*freq[k][w] for w in range(27)]
            ans.append(sum(count) - max(count))
        return ans
30600