Leetcode #2242: Maximum Score of a Node Sequence
In this guide, we solve Leetcode #2242 Maximum Score of a Node Sequence in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is an undirected graph with n nodes, numbered from 0 to n - 1. You are given a 0-indexed integer array scores of length n where scores[i] denotes the score of node i.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Array, Enumeration, Sorting
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: scores = [5,2,9,8,4], edges = [[0,1],[1,2],[2,3],[0,2],[1,3],[2,4]]
Output: 24
Explanation: The figure above shows the graph and the chosen node sequence [0,1,2,3].
The score of the node sequence is 5 + 2 + 9 + 8 = 24.
It can be shown that no other node sequence has a score of more than 24.
Note that the sequences [3,1,2,0] and [1,0,2,3] are also valid and have a score of 24.
The sequence [0,3,2,4] is not valid since no edge connects nodes 0 and 3.
Python Solution
class Solution:
def maximumScore(self, scores: List[int], edges: List[List[int]]) -> int:
g = defaultdict(list)
for a, b in edges:
g[a].append(b)
g[b].append(a)
for k in g.keys():
g[k] = nlargest(3, g[k], key=lambda x: scores[x])
ans = -1
for a, b in edges:
for c in g[a]:
for d in g[b]:
if b != c != d != a:
t = scores[a] + scores[b] + scores[c] + scores[d]
ans = max(ans, t)
return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.