Question

1014. Best Sightseeing Pair

中文文档

Description

You are given an integer array values where values[i] represents the value of the ith sightseeing spot. Two sightseeing spots i and j have a distance j - i between them.

The score of a pair (i < j) of sightseeing spots is values[i] + values[j] + i - j: the sum of the values of the sightseeing spots, minus the distance between them.

Return _the maximum score of a pair of sightseeing spots_.

 

Example 1:

Input: values = [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, values[i] + values[j] + i - j = 8 + 5 + 0 - 2 = 11

Example 2:

Input: values = [1,2]
Output: 2

 

Constraints:

• 2 <= values.length <= 5 * 104
• 1 <= values[i] <= 1000

Solutions

Solution 1

class Solution:
  def maxScoreSightseeingPair(self, values: List[int]) -> int:
    ans, mx = 0, values[0]
    for j in range(1, len(values)):
      ans = max(ans, values[j] - j + mx)
      mx = max(mx, values[j] + j)
    return ans

class Solution {
  public int maxScoreSightseeingPair(int[] values) {
    int ans = 0, mx = values[0];
    for (int j = 1; j < values.length; ++j) {
      ans = Math.max(ans, values[j] - j + mx);
      mx = Math.max(mx, values[j] + j);
    }
    return ans;
  }
}

class Solution {
public:
  int maxScoreSightseeingPair(vector<int>& values) {
    int ans = 0, mx = values[0];
    for (int j = 1; j < values.size(); ++j) {
      ans = max(ans, values[j] - j + mx);
      mx = max(mx, values[j] + j);
    }
    return ans;
  }
};

func maxScoreSightseeingPair(values []int) (ans int) {
  for j, mx := 1, values[0]; j < len(values); j++ {
    ans = max(ans, values[j]-j+mx)
    mx = max(mx, values[j]+j)
  }
  return
}

function maxScoreSightseeingPair(values: number[]): number {
  let ans = 0;
  let mx = values[0];
  for (let j = 1; j < values.length; ++j) {
    ans = Math.max(ans, values[j] - j + mx);
    mx = Math.max(mx, values[j] + j);
  }
  return ans;
}