<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 collections import defaultdict as dd
def find_lca(a,b,h,dp1,dp2):
    d1 = h[a]
    d2 = h[b]
    if d1 > d2:
        return find_lca(b,a,h,dp1,dp2)
    f = dd(int)
    diff = d2 - d1
    diff_bit = diff.bit_length()
    # print(diff,diff_bit)
    for v in range(diff_bit-1,-1,-1):
        if diff & (1 << v):
            d1 = dp2[b][v]
            b = dp1[b][v]
            for l,r in d1.items():
                f[l]+=r
    # print(f)
    if a==b:
        # print("equalx")
        total = 0
        c = 0
        for l,r in f.items():
            total+=r
            c = max(c,r)
        return total - c
    d_bit =  h[a].bit_length()
    # print(h[a],d_bit,"ssx")
    for v in range(d_bit-1,-1,-1):
        if dp1[a][v]!=-1 and dp1[a][v]!=dp1[b][v]:
            # print("indi ", v)
            d1 = dp2[a][v]
            d2 = dp2[b][v]
            for l,r in d1.items():
                f[l]+=r
            for l,r in d2.items():
                f[l]+=r
            a = dp1[a][v]
            b = dp1[b][v]
    d1 = dp2[a][0]
    d2 = dp2[b][0]
    for l,r in d1.items():
        f[l]+=r
    for l,r in d2.items():
        f[l]+=r
    total = 0
    c = 0
    # print(f)
    for l,r in f.items():
        total+=r
        c = max(c,r)
    # print(total,c)
    return total - c

class Solution:
    def minOperationsQueries(self, n: int, edges: List[List[int]], queries: List[List[int]]) -> List[int]:
        q = dd(list)
        weight = {}
        for u,v,w in edges:
            q[u].append(v)
            q[v].append(u)
            weight[(u,v)] = w
            weight[(v,u)] = w
        parent = {}
        visit = {0}
        h = {}
        h[0] = 0
        st = deque()
        st.append([0,0])
        v = n.bit_length()
        dp1 = [[-1 for j in range(v)] for i in range(n)]
        dp2 = [[dd(int) for j in range(v)] for i in range(n)]
        while st:
            a , d = st.popleft()
            for c in q[a]:
                if c not in visit:
                    st.append([c,d+1])
                    visit.add(c)
                    parent[c] = a
                    dp1[c][0] = a
                    cv = dp2[c][0]
                    cv[weight[(a,c)]]+=1
                    h[c] = d+1
        ans = []
        for j in range(1,v):
            for i in range(n):
                c = dp1[i][j-1]
                d1 = dp2[i][j-1]
                if c!=-1:
                    dp1[i][j] = dp1[c][j-1]
                    d2 = dp2[c][j-1]
                    e = dp2[i][j]
                    if dp1[i][j]!=-1:
                        for l,r in d1.items():
                            e[l]+=r
                        for l,r in d2.items():
                            e[l]+=r
        for a,b in queries:
            k = find_lca(a,b,h,dp1,dp2)
            ans.append(k)
        return ans
129300