LeetCode #3559 — HARD

Number of Ways to Assign Edge Weights II

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

Solve on LeetCode
The Problem

Problem Statement

There is an undirected tree with n nodes labeled from 1 to n, rooted at node 1. The tree is represented by a 2D integer array edges of length n - 1, where edges[i] = [ui, vi] indicates that there is an edge between nodes ui and vi.

Initially, all edges have a weight of 0. You must assign each edge a weight of either 1 or 2.

The cost of a path between any two nodes u and v is the total weight of all edges in the path connecting them.

You are given a 2D integer array queries. For each queries[i] = [ui, vi], determine the number of ways to assign weights to edges in the path such that the cost of the path between ui and vi is odd.

Return an array answer, where answer[i] is the number of valid assignments for queries[i].

Since the answer may be large, apply modulo 109 + 7 to each answer[i].

Note: For each query, disregard all edges not in the path between node ui and vi.

Example 1:

Input: edges = [[1,2]], queries = [[1,1],[1,2]]

Output: [0,1]

Explanation:

  • Query [1,1]: The path from Node 1 to itself consists of no edges, so the cost is 0. Thus, the number of valid assignments is 0.
  • Query [1,2]: The path from Node 1 to Node 2 consists of one edge (1 → 2). Assigning weight 1 makes the cost odd, while 2 makes it even. Thus, the number of valid assignments is 1.

Example 2:

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

Output: [2,1,4]

Explanation:

  • Query [1,4]: The path from Node 1 to Node 4 consists of two edges (1 → 3 and 3 → 4). Assigning weights (1,2) or (2,1) results in an odd cost. Thus, the number of valid assignments is 2.
  • Query [3,4]: The path from Node 3 to Node 4 consists of one edge (3 → 4). Assigning weight 1 makes the cost odd, while 2 makes it even. Thus, the number of valid assignments is 1.
  • Query [2,5]: The path from Node 2 to Node 5 consists of three edges (2 → 1, 1 → 3, and 3 → 5). Assigning (1,2,2), (2,1,2), (2,2,1), or (1,1,1) makes the cost odd. Thus, the number of valid assignments is 4.

Constraints:

  • 2 <= n <= 105
  • edges.length == n - 1
  • edges[i] == [ui, vi]
  • 1 <= queries.length <= 105
  • queries[i] == [ui, vi]
  • 1 <= ui, vi <= n
  • 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: There is an undirected tree with n nodes labeled from 1 to n, rooted at node 1. The tree is represented by a 2D integer array edges of length n - 1, where edges[i] = [ui, vi] indicates that there is an edge between nodes ui and vi. Initially, all edges have a weight of 0. You must assign each edge a weight of either 1 or 2. The cost of a path between any two nodes u and v is the total weight of all edges in the path connecting them. You are given a 2D integer array queries. For each queries[i] = [ui, vi], determine the number of ways to assign weights to edges in the path such that the cost of the path between ui and vi is odd. Return an array answer, where answer[i] is the number of valid assignments for queries[i]. Since the answer may be large, apply modulo 109 + 7 to each answer[i]. Note: For each query, disregard all edges not in the path between node ui and vi.

Baseline thinking

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

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

Example 1

[[1,2]]
[[1,1],[1,2]]

Example 2

[[1,2],[1,3],[3,4],[3,5]]
[[1,4],[3,4],[2,5]]
Step 02

Core Insight

What unlocks the optimal approach

  • Dynamic programming with states <code>chainLength</code> and <code>sumParity</code>.
  • Use Lowest Common Ancestor to find the distance between any two nodes quickly in <code>O(logn)</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 #3559: Number of Ways to Assign Edge Weights II
class Solution {
  public int[] assignEdgeWeights(int[][] edges, int[][] queries) {
    final int n = edges.length + 1;
    int[] ans = new int[queries.length];
    final int[] depth = new int[n + 1];
    final int[][] parent = new int[LOG][n + 1];
    List<Integer>[] graph = new List[n + 1];
    Arrays.setAll(graph, i -> new ArrayList<>());

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      graph[u].add(v);
      graph[v].add(u);
    }

    dfs(1, -1, graph, parent, depth);

    for (int k = 1; k < LOG; ++k)
      for (int v = 1; v <= n; ++v)
        if (parent[k - 1][v] != -1)
          parent[k][v] = parent[k - 1][parent[k - 1][v]];

    for (int i = 0; i < queries.length; ++i) {
      final int u = queries[i][0];
      final int v = queries[i][1];
      if (u == v) {
        ans[i] = 0;
      } else {
        final int a = lca(u, v, parent, depth);
        final int d = depth[u] + depth[v] - 2 * depth[a];
        ans[i] = modPow(2, d - 1);
      }
    }

    return ans;
  }

  private static final int MOD = 1_000_000_007;
  private static final int LOG = 17; // since 2^17 > 1e5

  private void dfs(int u, int p, java.util.List<Integer>[] graph, int[][] parent, int[] depth) {
    parent[0][u] = p;
    for (int v : graph[u]) {
      if (v != p) {
        depth[v] = depth[u] + 1;
        dfs(v, u, graph, parent, depth);
      }
    }
  }

  private int lca(int u, int v, int[][] parent, int[] depth) {
    if (depth[u] < depth[v]) {
      final int temp = u;
      u = v;
      v = temp;
    }

    for (int k = LOG - 1; k >= 0; --k)
      if (parent[k][u] != -1 && depth[parent[k][u]] >= depth[v])
        u = parent[k][u];

    if (u == v)
      return u;

    for (int k = LOG - 1; k >= 0; --k)
      if (parent[k][u] != -1 && parent[k][u] != parent[k][v]) {
        u = parent[k][u];
        v = parent[k][v];
      }

    return parent[0][u];
  }

  private int modPow(long x, long n) {
    if (n == 0)
      return 1;
    if (n % 2 == 1)
      return (int) (x * modPow(x % MOD, (n - 1)) % MOD);
    return modPow(x * x % MOD, (n / 2)) % MOD;
  }
}
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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.

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