How do you solve Number of Ways to Assign Edge Weights I on LeetCode?

Number of Ways to Assign Edge Weights I (LeetCode #3558) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The path to the deepest node determines the edge count.

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

The optimal solution for Number of Ways to Assign Edge Weights I runs in O(n) time and uses O(n) space. DFS traverses each node once, leading to linear complexity relative to the number of nodes.

What is the brute force solution for Number of Ways to Assign Edge Weights I?

The brute force approach for Number of Ways to Assign Edge Weights I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible combinations of edge weights (1 or 2) for the path from node 1 to the deepest node. Count those combinations that yield an odd total weight. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Ways to Assign Edge Weights I?

Number of Ways to Assign Edge Weights I uses the following concepts: Math, Tree, Depth-First Search. The recommended patterns to study are: Depth-First Search, Dynamic Programming.

What are the interview tips for Number of Ways to Assign Edge Weights I?

Explain your thought process clearly. Use examples to illustrate your approach. Practice DFS and tree traversal problems.

What are common mistakes when solving Number of Ways to Assign Edge Weights I?

Forgetting to consider modulo operation. Not handling edge cases like single-node trees.

#3558

Number of Ways to Assign Edge Weights I

Medium

MathTreeDepth-First SearchDepth-First SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Use DFS to find the maximum depth and count the edges in the path. The total weight can be odd if the number of edges with weight 1 is odd.

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Algorithm

3 steps

1Step 1: Perform DFS to find the maximum depth and count edges in the path to that depth.
2Step 2: Calculate the number of ways to assign weights such that the total weight is odd.
3Step 3: Return the result modulo 10^9 + 7.

solution.py18 lines

1def count_odd_assignments(edges):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for u, v in edges:
5        graph[u].append(v)
6        graph[v].append(u)
7    max_depth = 0
8    def dfs(node, parent, depth):
9        nonlocal max_depth
10        max_depth = max(max_depth, depth)
11        for neighbor in graph[node]:
12            if neighbor != parent:
13                dfs(neighbor, node, depth + 1)
14    dfs(1, -1, 0)
15    return pow(2, max_depth - 1, 10**9 + 7) if max_depth > 0 else 0
16
17# Example usage
18print(count_odd_assignments([[1, 2]]))

ℹ

Complexity note: DFS traverses each node once, leading to linear complexity relative to the number of nodes.

1The path to the deepest node determines the edge count.
2Odd total weights require careful assignment of weights.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.