How do you solve Maximum Score of a Node Sequence on LeetCode?

Maximum Score of a Node Sequence (LeetCode #2242) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + e log e) time complexity and O(n + e) space complexity. Key insight: Fixing two nodes connected by an edge significantly reduces the search space.

What is the brute force solution for Maximum Score of a Node Sequence?

The brute force approach for Maximum Score of a Node Sequence is Brute Force, with O(n^4) time complexity and O(n) space complexity. The brute force approach involves generating all possible sequences of length 4 from the nodes and checking if they are valid based on the edges. This method is straightforward but inefficient for larger graphs. The optimal Optimal Solution approach improves this to O(n + e log e).

What algorithm or data structure is used to solve Maximum Score of a Node Sequence?

Maximum Score of a Node Sequence uses the following concepts: Array, Graph Theory, Sorting, Enumeration. The recommended patterns to study are: Graph Traversal, Sorting.

What are the interview tips for Maximum Score of a Node Sequence?

Always clarify the problem constraints and edge cases. Think about how to reduce the problem size early on. Practice explaining your thought process clearly during coding.

What are common mistakes when solving Maximum Score of a Node Sequence?

Not considering all valid sequences when checking edges. Overlooking the need to sort candidates to find the best scores.

#2242

Maximum Score of a Node Sequence

Hard

ArrayGraph TheorySortingEnumerationGraph TraversalSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^4)	O(n + e log e)
Space	O(n)	O(n + e)

💡

Intuition

Time O(n + e log e)Space O(n + e)

The optimal solution leverages the idea of fixing two nodes connected by an edge and searching for the best two additional nodes to maximize the score. This reduces the number of checks significantly compared to the brute force method.

⚙️

Algorithm

4 steps

1Step 1: Create an adjacency list from the edges to represent the graph.
2Step 2: For each edge (u, v), consider u and v as the middle nodes of the sequence.
3Step 3: For each of the two middle nodes, find the top two nodes connected to them that maximize the score.
4Step 4: Calculate the score for the sequence formed by u, v, and the two best nodes found, and keep track of the maximum score.

solution.py23 lines

1# Full working Python code
2from collections import defaultdict
3
4def maxScore(scores, edges):
5    n = len(scores)
6    graph = defaultdict(list)
7    for u, v in edges:
8        graph[u].append(v)
9        graph[v].append(u)
10    max_score = -1
11    for u, v in edges:
12        candidates = []
13        for neighbor in graph[u]:
14            if neighbor != v:
15                candidates.append(scores[neighbor])
16        for neighbor in graph[v]:
17            if neighbor != u:
18                candidates.append(scores[neighbor])
19        candidates.sort(reverse=True)
20        if len(candidates) >= 2:
21            score = scores[u] + scores[v] + candidates[0] + candidates[1]
22            max_score = max(max_score, score)
23    return max_score

ℹ

Complexity note: The time complexity is O(n + e log e) due to building the adjacency list and sorting the candidates. The space complexity is O(n + e) for storing the graph and candidates.

1Fixing two nodes connected by an edge significantly reduces the search space.
2Using an adjacency list allows efficient access to connected nodes.

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