LeetCode #2242 — HARD

Maximum Score of a Node Sequence

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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. You are also given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting nodes ai and bi.

A node sequence is valid if it meets the following conditions:

  • There is an edge connecting every pair of adjacent nodes in the sequence.
  • No node appears more than once in the sequence.

The score of a node sequence is defined as the sum of the scores of the nodes in the sequence.

Return the maximum score of a valid node sequence with a length of 4. If no such sequence exists, return -1.

Example 1:

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.

Example 2:

Input: scores = [9,20,6,4,11,12], edges = [[0,3],[5,3],[2,4],[1,3]]
Output: -1
Explanation: The figure above shows the graph.
There are no valid node sequences of length 4, so we return -1.

Constraints:

  • n == scores.length
  • 4 <= n <= 5 * 104
  • 1 <= scores[i] <= 108
  • 0 <= edges.length <= 5 * 104
  • edges[i].length == 2
  • 0 <= ai, bi <= n - 1
  • ai != bi
  • There are no duplicate edges.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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. You are also given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting nodes ai and bi. A node sequence is valid if it meets the following conditions: There is an edge connecting every pair of adjacent nodes in the sequence. No node appears more than once in the sequence. The score of a node sequence is defined as the sum of the scores of the nodes in the sequence. Return the maximum score of a valid node sequence with a length of 4. If no such sequence exists, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[5,2,9,8,4]
[[0,1],[1,2],[2,3],[0,2],[1,3],[2,4]]

Example 2

[9,20,6,4,11,12]
[[0,3],[5,3],[2,4],[1,3]]

Related Problems

  • Get the Maximum Score (get-the-maximum-score)
Step 02

Core Insight

What unlocks the optimal approach

  • For every node sequence of length 4, there are 3 relevant edges. How can we consider valid triplets of edges?
  • Fix the middle 2 nodes connected by an edge in the node sequence. Can you determine the other 2 nodes that will give the highest possible score?
  • The other 2 nodes must each be connected to one of the middle nodes. If we only consider nodes with the highest scores, how many should we store to ensure we don’t choose duplicate nodes?
  • For each node, we should store the 3 adjacent nodes with the highest scores to ensure we can find a sequence with no duplicate nodes via the method above.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2242: Maximum Score of a Node Sequence
class Solution {
    public int maximumScore(int[] scores, int[][] edges) {
        int n = scores.length;
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        for (int i = 0; i < n; ++i) {
            g[i].sort((a, b) -> scores[b] - scores[a]);
            g[i] = g[i].subList(0, Math.min(3, g[i].size()));
        }
        int ans = -1;
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            for (int c : g[a]) {
                for (int d : g[b]) {
                    if (c != b && c != d && a != d) {
                        int t = scores[a] + scores[b] + scores[c] + scores[d];
                        ans = Math.max(ans, t);
                    }
                }
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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