Problem: Minimum Edge Weight Equilibrium Queries in a Tree

There is an undirected tree with `n` nodes labeled from `0` to `n - 1`. You
are given the integer `n` and a 2D integer array `edges` of length `n - 1`,
where `edges[i] = [ui, vi, wi]` indicates that there is an edge between nodes
`ui` and `vi` with weight `wi` in the tree.

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

Note that:

* Queries are independent of each other, meaning that the tree returns to its initial state on each new query.
* The path from `ai` to `bi` is a sequence of distinct nodes starting with node `ai` and ending with node `bi` such that every two adjacent nodes in the sequence share an edge in the tree.

Return an array `answer` of length `m` where `answer[i]` is the answer to the
`ith` query.


Python solution:
import math
import itertools
import bisect
import collections
import string
import heapq
import functools
import sortedcontainers
from typing import List, Dict, Tuple, Iterator

class Solution:
  def minOperationsQueries(self, n: int, edges: List[List[int]], queries: List[List[int]]) -> List[int]:
    kMax = 26
    m = int(math.log2(n)) + 1
    ans = []
    graph = [[] for _ in range(n)]
    jump = [[0] * m for _ in range(n)]
    count = [[] for _ in range(n)]
    depth = [0] * n

    for u, v, w in edges:
      graph[u].append((v, w))
      graph[v].append((u, w))

    def dfs(u: int, prev: int, d: int):
      if prev != -1:
        jump[u][0] = prev
      depth[u] = d
      for v, w in graph[u]:
        if v == prev:
          continue
        count[v] = count[u][:]
        count[v][w] += 1
        dfs(v, u, d + 1)

    count[0] = [0] * (kMax + 1)
    dfs(0, -1, 0)

    for j in range(1, m):
      for i in range(n):
        jump[i][j] = jump[jump[i][j - 1]][j - 1]

    def getLCA(u: int, v: int) -> int:
      if depth[u] > depth[v]:
        return getLCA(v, u)
      for j in range(m):
        if depth[v] - depth[u] >> j & 1:
          v = jump[v][j]
      if u == v:
        return u
      for j in range(m - 1, -1, -1):
        if jump[u][j] != jump[v][j]:
          u = jump[u][j]
          v = jump[v][j]
      return jump[v][0]

    for u, v in queries:
      lca = getLCA(u, v)
      numEdges = depth[u] + depth[v] - 2 * depth[lca]
      maxFreq = max(count[u][j] + count[v][j] - 2 * count[lca][j] for j in range(1, kMax + 1))
      ans.append(numEdges - maxFreq)

    return ans


Java solution:
class Solution {
  public int[] minOperationsQueries(int n, int[][] edges, int[][] queries) {
    final int MAX = 26;
    final int m = (int) Math.ceil(Math.log(n) / Math.log(2));
    int[] ans = new int[queries.length];
    List<Pair<Integer, Integer>>[] graph = new List[n];
    // jump[i][j] := the 2^j-th ancestor of i
    int[][] jump = new int[n][m];
    // depth[i] := the depth of i
    int[] depth = new int[n];
    // count[i][j] := the count of j from root to i, where 1 <= j <= 26
    int[][] count = new int[n][MAX + 1];
    Arrays.setAll(graph, i -> new ArrayList<>());

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      final int w = edge[2];
      graph[u].add(new Pair<>(v, w));
      graph[v].add(new Pair<>(u, w));
    }

    count[0] = new int[MAX + 1];
    dfs(graph, 0, /*prev=*/-1, jump, depth, count);

    for (int j = 1; j < m; ++j)
      for (int i = 0; i < n; ++i)
        jump[i][j] = jump[jump[i][j - 1]][j - 1];

    for (int i = 0; i < queries.length; ++i) {
      final int u = queries[i][0];
      final int v = queries[i][1];
      final int lca = getLCA(u, v, jump, depth);
      // the number of edges between (u, v).
      final int numEdges = depth[u] + depth[v] - 2 * depth[lca];
      // the maximum frequency of edges between (u, v)
      int maxFreq = 0;
      for (int j = 1; j <= MAX; ++j)
        maxFreq = Math.max(maxFreq, count[u][j] + count[v][j] - 2 * count[lca][j]);
      ans[i] = numEdges - maxFreq;
    }

    return ans;
  }

  private void dfs(List<Pair<Integer, Integer>>[] graph, int u, int prev, int[][] jump, int[] depth,
                   int[][] count) {
    for (Pair<Integer, Integer> pair : graph[u]) {
      final int v = pair.getKey();
      final int w = pair.getValue();
      if (v == prev)
        continue;
      jump[v][0] = u;
      depth[v] = depth[u] + 1;
      count[v] = count[u].clone();
      ++count[v][w];
      dfs(graph, v, u, jump, depth, count);
    }
  }

  // Returns the lca(u, v) by binary jump.
  private int getLCA(int u, int v, int[][] jump, int[] depth) {
    // v is always deeper than u.
    if (depth[u] > depth[v])
      return getLCA(v, u, jump, depth);
    // Jump v to the same height of u.
    for (int j = 0; j < jump[0].length; ++j)
      if ((depth[v] - depth[u] >> j & 1) == 1)
        v = jump[v][j];
    if (u == v)
      return u;
    // Jump u and v to the node right below the lca.
    for (int j = jump[0].length - 1; j >= 0; --j)
      if (jump[u][j] != jump[v][j]) {
        u = jump[u][j];
        v = jump[v][j];
      }
    return jump[v][0];
  }
}


C++ solution:
class Solution {
 public:
  vector<int> minOperationsQueries(int n, vector<vector<int>>& edges,
                                   vector<vector<int>>& queries) {
    constexpr int kMax = 26;
    const int m = ceil(log2(n));
    vector<int> ans;
    vector<vector<pair<int, int>>> graph(n);
    // jump[i][j] := the 2^j-th ancestor of i
    vector<vector<int>> jump(n, vector<int>(m));
    // depth[i] := the depth of i
    vector<int> depth(n);
    // count[i][j] := the count of j from root to i, where 1 <= j <= 26
    vector<vector<int>> count(n);

    for (const vector<int>& edge : edges) {
      const int u = edge[0];
      const int v = edge[1];
      const int w = edge[2];
      graph[u].emplace_back(v, w);
      graph[v].emplace_back(u, w);
    }

    count[0] = vector<int>(kMax + 1);
    dfs(graph, 0, /*prev=*/-1, jump, depth, count);

    for (int j = 1; j < m; ++j)
      for (int i = 0; i < n; ++i)
        jump[i][j] = jump[jump[i][j - 1]][j - 1];

    for (const vector<int>& query : queries) {
      const int u = query[0];
      const int v = query[1];
      const int lca = getLCA(u, v, jump, depth);
      // the number of edges between (u, v).
      const int numEdges = depth[u] + depth[v] - 2 * depth[lca];
      // the maximum frequency of edges between (u, v)
      int maxFreq = 0;
      for (int j = 1; j <= kMax; ++j)
        maxFreq = max(maxFreq, count[u][j] + count[v][j] - 2 * count[lca][j]);
      ans.push_back(numEdges - maxFreq);
    }

    return ans;
  }

 private:
  void dfs(const vector<vector<pair<int, int>>>& graph, int u, int prev,
           vector<vector<int>>& jump, vector<int>& depth,
           vector<vector<int>>& count) {
    for (const auto& [v, w] : graph[u]) {
      if (v == prev)
        continue;
      jump[v][0] = u;
      depth[v] = depth[u] + 1;
      count[v] = count[u];
      ++count[v][w];
      dfs(graph, v, u, jump, depth, count);
    }
  }

  // Returns the lca(u, v) by binary jump.
  int getLCA(int u, int v, const vector<vector<int>>& jump,
             const vector<int>& depth) {
    // v is always deeper than u.
    if (depth[u] > depth[v])
      return getLCA(v, u, jump, depth);
    // Jump v to the same height of u.
    for (int j = 0; j < jump[0].size(); ++j)
      if (depth[v] - depth[u] >> j & 1)
        v = jump[v][j];
    if (u == v)
      return u;
    // Jump u and v to the node right below the lca.
    for (int j = jump[0].size() - 1; j >= 0; --j)
      if (jump[u][j] != jump[v][j]) {
        u = jump[u][j];
        v = jump[v][j];
      }
    return jump[v][0];
  }
};