Is Number of Ways to Assign Edge Weights II easy or hard?

Number of Ways to Assign Edge Weights II is classified as Hard. The challenge comes from combining tree traversal with dynamic programming and efficiently representing constraint states using bit manipulation or compressed DP states.

Number of Ways to Assign Edge Weights II Python/Java solution

Implement the algorithm by building an adjacency list for the tree and running a DFS from the root. During traversal, maintain DP arrays or maps that store the number of valid configurations for each state. The same logic translates cleanly across Python, Java, C++, and Go.

How to solve Number of Ways to Assign Edge Weights II in O(n)?

A near-linear solution is achieved by performing a single DFS traversal of the tree and computing DP states at each node. Instead of enumerating all assignments, the algorithm aggregates counts from child subtrees and updates states using constant-time transitions. This keeps the traversal close to O(n) when the number of states is small.

What is the best approach for Number of Ways to Assign Edge Weights II?

The most efficient method uses tree dynamic programming combined with depth-first search. Each node computes the number of valid configurations for its subtree based on the state passed from its parent edge. Bitmasking or compact state encoding keeps transitions fast, leading to an overall complexity of O(n * states).

Is Number of Ways to Assign Edge Weights II asked at Google/Amazon/Meta?

Problems involving tree DP, edge assignments, and state propagation appear frequently in interviews at companies like Google, Amazon, and Meta. Variants of this problem test understanding of DFS traversal, subtree aggregation, and dynamic programming on trees.

What data structure is used in Number of Ways to Assign Edge Weights II?

The core structure is a tree represented with an adjacency list. The solution relies on depth-first search to traverse nodes and dynamic programming tables to store valid assignment counts for each state. Bit manipulation is often used to encode constraints efficiently.

What is the time complexity of Number of Ways to Assign Edge Weights II?

The optimal tree DP solution runs in O(n * s) time, where n is the number of nodes and s is the number of possible constraint states represented in the DP. Space complexity is also O(n * s) because each node stores results for its DP states.

Description Hints 2 Solution Video 38:53

3559. Number of Ways to Assign Edge Weights II

Hard71.8% acceptance3 companies

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] = [u_i, v_i] indicates that there is an edge between nodes u_i and v_i.

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] = [u_i, v_i], determine the number of ways to assign weights to edges in the path such that the cost of the path between u_i and v_i 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 u_i and v_i.

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] == [u_i, v_i]
1 <= queries.length <= 10⁵
queries[i] == [u_i, v_i]
1 <= u_i, v_i <= n
edges represents a valid tree.

Asked at 3 companies

Bloomberg

Google Amazon

Topics

Hints open one at a time

Hint 1

Dynamic programming with states chainLength and sumParity.

Hint 2

Use Lowest Common Ancestor to find the distance between any two nodes quickly in O(logn).

38:53

Still stuck?Read the written solution, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
2    def assignEdgeWeights(self, edges: List[List[int]], queries: List[List[int]]) -> List[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

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

Input

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

Your code is saved on this device.Solution