Question

2242. Maximum Score of a Node Sequence

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Description

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.

Solutions

Solution 1

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

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;
  }
}