<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]:
        # node_weight[node][weight] = freq
        node_weight_freq = defaultdict(lambda: defaultdict(lambda : 0))
        # up[node][i] = 2^i th ancestor_node for current node
        up = defaultdict(list)
        # depth[node] = the depth of current node
        depth = defaultdict(lambda:0)

        G = defaultdict(list)
        for u,v,w in edges:
            G[u].append([v,w])
            G[v].append([u,w])

        def dfs(i, seen, weightCnt, path):
            for w, freq in weightCnt.items():
                node_weight_freq[i][w] = freq
            depth[i] = len(path) - 1
            
            if depth[i] > 0:
                for j in range(0, ceil(log2(depth[i]))):
                    up[i].append(path[-2**j-1])
            up[i].append(path[0])

            for to, w in G[i]:
                if to in seen: continue
                seen.add(to)

                path.append(to)
                weightCnt[w] += 1
                dfs(to, seen, weightCnt, path)
                path.pop()
                weightCnt[w] -= 1

        dfs(0, set([0]), defaultdict(lambda: 0), [0])

        @cache
        def lca(u, v) -> int:
            # u alway points to the higher node
            if depth[u] > depth[v]: u,v = v, u
            if u == 0: return u
            # check if u is the ancesstor of v
            while depth[v] != depth[u]:
                depth_diff = depth[v] - depth[u]
                v = up[v][floor(log2(depth_diff))]
            if v == u: return u
            
            # u and v are on the same depth now, need to find their common ancesstor
            while up[u][0] != up[v][0]:   
                max_diff = ceil(log2(depth[v]))+1
                for i in range(0, max_diff):
                    if up[u][i] != up[v][i]:
                        u = up[u][i]
                        v = up[v][i]
                        break

            return up[u][0]

        @cache
        def getAns(u,v):
            ancestor = lca(u,v)
            weight_freq = [node_weight_freq[u][i] + node_weight_freq[v][i] - 2*node_weight_freq[ancestor][i] for i in range(1,27)]
            return sum(weight_freq)-max(weight_freq)
        

        ans = []
        for u,v in queries:
            weight_freq = [0] * 27
            u,v = min(u,v), max(u,v)
            ans.append(getAns(u,v))

        return ans
50000