LeetCode #3553 — HARD

Minimum Weighted Subgraph With the Required Paths II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an undirected weighted tree with n nodes, numbered from 0 to n - 1. It is represented by 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.​

Additionally, you are given a 2D integer array queries, where queries[j] = [src1j, src2j, destj].

Return an array answer of length equal to queries.length, where answer[j] is the minimum total weight of a subtree such that it is possible to reach destj from both src1j and src2j using edges in this subtree.

A subtree here is any connected subset of nodes and edges of the original tree forming a valid tree.

Example 1:

Input: edges = [[0,1,2],[1,2,3],[1,3,5],[1,4,4],[2,5,6]], queries = [[2,3,4],[0,2,5]]

Output: [12,11]

Explanation:

The blue edges represent one of the subtrees that yield the optimal answer.

  • answer[0]: The total weight of the selected subtree that ensures a path from src1 = 2 and src2 = 3 to dest = 4 is 3 + 5 + 4 = 12.

  • answer[1]: The total weight of the selected subtree that ensures a path from src1 = 0 and src2 = 2 to dest = 5 is 2 + 3 + 6 = 11.

Example 2:

Input: edges = [[1,0,8],[0,2,7]], queries = [[0,1,2]]

Output: [15]

Explanation:

  • answer[0]: The total weight of the selected subtree that ensures a path from src1 = 0 and src2 = 1 to dest = 2 is 8 + 7 = 15.

Constraints:

  • 3 <= n <= 105
  • edges.length == n - 1
  • edges[i].length == 3
  • 0 <= ui, vi < n
  • 1 <= wi <= 104
  • 1 <= queries.length <= 105
  • queries[j].length == 3
  • 0 <= src1j, src2j, destj < n
  • src1j, src2j, and destj are pairwise distinct.
  • The input is generated such that edges represents a valid tree.
Patterns Used
📊Dynamic Programming→⚡Bit Manipulation→🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an undirected weighted tree with n nodes, numbered from 0 to n - 1. It is represented by 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.​ Additionally, you are given a 2D integer array queries, where queries[j] = [src1j, src2j, destj]. Return an array answer of length equal to queries.length, where answer[j] is the minimum total weight of a subtree such that it is possible to reach destj from both src1j and src2j using edges in this subtree. A subtree here is any connected subset of nodes and edges of the original tree forming a valid tree.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Bit Manipulation · Tree

Example 1

[[0,1,2],[1,2,3],[1,3,5],[1,4,4],[2,5,6]]
[[2,3,4],[0,2,5]]

Example 2

[[1,0,8],[0,2,7]]
[[0,1,2]]
Step 02

Core Insight

What unlocks the optimal approach

  • Binary lifting
  • Find the lowest common ancestor (LCA) of any two nodes using binary lifting
  • For any node <code>x</code>, let <code>f(x)</code> be the distance from the root to <code>x</code>. Then for two nodes <code>x</code> and <code>y</code>:<code>d(x, y) = f(x) + f(y) - 2 * f(LCA(x, y))</code>
  • For three nodes <code>a</code>, <code>b</code> and <code>c</code>, the minimum total weight of the subtree connecting all three is:<code>(d(a, b) + d(b, c) + d(c, a)) / 2</code>, where <code>d(x, y)</code> is the distance between nodes <code>x</code> and <code>y</code>
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3553: Minimum Weighted Subgraph With the Required Paths II
class Solution {
  // Similar to 2846. Minimum Edge Weight Equilibrium Queries in a Tree
  public int[] minimumWeight(int[][] edges, int[][] queries) {
    final int n = edges.length + 1;
    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];
    // dist[i] := the distance from root to i
    int[] dist = new int[n];
    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));
    }

    dfs(graph, 0, /*prev=*/-1, jump, depth, dist);

    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 src1 = queries[i][0];
      final int src2 = queries[i][1];
      final int dest = queries[i][2];
      ans[i] = (distance(src1, src2, jump, depth, dist) + distance(src1, dest, jump, depth, dist) +
                distance(src2, dest, jump, depth, dist)) /
               2;
    }

    return ans;
  }

  private void dfs(List<Pair<Integer, Integer>>[] graph, int u, int prev, int[][] jump, int[] depth,
                   int[] dist) {
    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;
      dist[v] = dist[u] + w;
      dfs(graph, v, u, jump, depth, dist);
    }
  }

  // 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[u][0];
  }

  // Returns the distance between u and v.
  private int distance(int u, int v, int[][] jump, int[] depth, int[] dist) {
    final int lca = getLCA(u, v, jump, depth);
    return dist[u] + dist[v] - 2 * dist[lca];
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.

Related Problems
#2920Maximum Points After Collecting Coins From All Nodeshard→#3068Find the Maximum Sum of Node Valueshard→#3559Number of Ways to Assign Edge Weights IIhard→#3575Maximum Good Subtree Scorehard→#3585Find Weighted Median Node in Treehard→