Introduction

In this blog post, we will explore how to solve the LeetCode problem "2242" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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.

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Iterate through each edge in the graph, considering the two nodes connected by the edge as the middle two nodes of a potential length-4 sequence.
  • For each such edge, find the top 3 neighbors of each of the two middle nodes (excluding the other middle node to avoid duplicates).
  • Calculate the score of all possible combinations of these neighbors and the middle nodes, and maintain the maximum score found so far.

• Complexity:

  • Runtime: O(E + N), where E is the number of edges and N is the number of nodes. The dominant factor is iterating through the edges, and then the constant factor of searching top 3 neighbors for each node.
  • Storage: O(N + E). Storing the adjacency list and neighbor lists takes up space.

Code

    def max_score_of_a_node_sequence(scores, edges):
    n = len(scores)
    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)

    max_score = -1

    for u, v in edges:
        neighbors_u = []
        for neighbor in adj[u]:
            if neighbor != v:
                neighbors_u.append((scores[neighbor], neighbor))
        neighbors_u.sort(reverse=True)
        neighbors_u = neighbors_u[:3]

        neighbors_v = []
        for neighbor in adj[v]:
            if neighbor != u:
                neighbors_v.append((scores[neighbor], neighbor))
        neighbors_v.sort(reverse=True)
        neighbors_v = neighbors_v[:3]

        for score_u, node_u in neighbors_u:
            for score_v, node_v in neighbors_v:
                if node_u != node_v:
                    current_score = scores[u] + scores[v] + score_u + score_v
                    max_score = max(max_score, current_score)

    return max_score