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Problem solution · Java

Minimum Edge Weight Equilibrium Queries in a Tree

Minimum Edge Weight Equilibrium Queries in a Tree: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
81 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Minimum Edge Weight Equilibrium Queries in a Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 81 lines of Java from the credited upstream file 2846.java.
  • The implementation visibly relies on sequence storage.
  • 8 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Edge Weight Equilibrium Queries in a Tree · JavaJava
Use this to learn the idea, then write your own version.
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];  }}

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