How do you solve Maximum Subgraph Score in a Tree on LeetCode?

Maximum Subgraph Score in a Tree (LeetCode #3772) 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: Rerooting allows efficient score adjustment without redundant calculations.

What is the time complexity of Maximum Subgraph Score in a Tree?

The optimal solution for Maximum Subgraph Score in a Tree runs in O(n) time and uses O(n) space. Each node is visited a constant number of times during DFS and rerooting, leading to O(n).

What is the brute force solution for Maximum Subgraph Score in a Tree?

The brute force approach for Maximum Subgraph Score in a Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check all possible connected subgraphs for each node to find the maximum score. This involves exploring all combinations, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Subgraph Score in a Tree?

Maximum Subgraph Score in a Tree uses the following concepts: Array, Dynamic Programming, Tree, Depth-First Search. The recommended patterns to study are: Dynamic Programming, Tree Traversal.

What are the interview tips for Maximum Subgraph Score in a Tree?

Explain your thought process clearly. Discuss time and space complexities. Practice similar tree problems.

What are common mistakes when solving Maximum Subgraph Score in a Tree?

Not considering the parent node during DFS. Failing to reset visited nodes in rerooting.

#3772

Maximum Subgraph Score in a Tree

Hard

ArrayDynamic ProgrammingTreeDepth-First SearchDynamic ProgrammingTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

We can use a rerooting technique with DFS to calculate scores efficiently. By calculating scores for subtrees and adjusting when rerooting, we avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Perform a DFS to calculate the initial score for each subtree.
2Step 2: Use rerooting to adjust scores for each node based on its parent's score.
3Step 3: Store the maximum score for each node.

solution.py23 lines

1def maxScoreOptimal(n, edges, good):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for a, b in edges:
5        graph[a].append(b)
6        graph[b].append(a)
7    scores = [0] * n
8    def dfs(node, parent):
9        score = good[node]
10        for neighbor in graph[node]:
11            if neighbor != parent:
12                score += dfs(neighbor, node)
13        scores[node] = score
14        return score
15    dfs(0, -1)
16    result = [0] * n
17    def reroot(node, parent, parentScore):
18        result[node] = scores[node] + parentScore
19        for neighbor in graph[node]:
20            if neighbor != parent:
21                reroot(neighbor, node, result[node] - scores[neighbor])
22    reroot(0, -1, 0)
23    return result

ℹ

Complexity note: Each node is visited a constant number of times during DFS and rerooting, leading to O(n).

1Rerooting allows efficient score adjustment without redundant calculations.
2DFS helps in calculating subtree scores effectively.

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