LeetCode #3558 — MEDIUM

Number of Ways to Assign Edge Weights I

Move from brute-force thinking to an efficient approach using math strategy.

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.

Select any one node x at the maximum depth. Return the number of ways to assign edge weights in the path from node 1 to x such that its total cost is odd.

Since the answer may be large, return it modulo 109 + 7.

Note: Ignore all edges not in the path from node 1 to x.

Example 1:

Input: edges = [[1,2]]

Output: 1

Explanation:

  • 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]]

Output: 2

Explanation:

  • The maximum depth is 2, with nodes 4 and 5 at the same depth. Either node can be selected for processing.
  • For example, 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.

Constraints:

  • 2 <= n <= 105
  • edges.length == n - 1
  • edges[i] == [ui, vi]
  • 1 <= ui, vi <= n
  • edges represents a valid tree.
Patterns Used
🌳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. Select any one node x at the maximum depth. Return the number of ways to assign edge weights in the path from node 1 to x such that its total cost is odd. Since the answer may be large, return it modulo 109 + 7. Note: Ignore all edges not in the path from node 1 to x.

Baseline thinking

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

Pattern signal: Math · Tree

Example 1

[[1,2]]

Example 2

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

Core Insight

What unlocks the optimal approach

  • Depth‑First Search (DFS) to compute the depth of each node from the root.
  • Find the maximum depth, <code>max_depth</code>.
  • The number of <code>2</code>s doesn’t affect parity; we only need an odd number of <code>1</code>s along the path.
  • The number of ways to choose an odd count of 1s among <code>max_depth</code> edges is <code>2^(max_depth-1)</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
Upper-end input sizes
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 #3558: Number of Ways to Assign Edge Weights I
class Solution {
  public int assignEdgeWeights(int[][] edges) {
    final int n = edges.size() + 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);
    }

    Queue<Integer> q = new ArrayDeque<>(List.of(1));
    boolean[] seen = new boolean[n + 1];
    seen[1] = true;

    int step = 0;
    for (step = 0; !q.isEmpty(); ++step)
      for (int sz = q.size(); sz > 0; --sz) {
        final int u = q.poll();
        for (final int v : graph[u])
          if (!seen[v]) {
            q.offer(v);
            seen[v] = true;
          }
      }

    return step > 0 ? modPow(2, step - 2) : 0;
  }

  private static final int MOD = 1_000_000_007;

  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)
Space
O(h)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

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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