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1014. Best Sightseeing Pair

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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  };
  

• 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
  
  

• 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;
  }
  
  

• impl Solution {
      pub fn max_score_sightseeing_pair(values: Vec<i32>) -> i32 {
          let mut ans = 0;
          let mut mx = 0;
          for (j, &x) in values.iter().enumerate() {
              ans = ans.max(mx + x - j as i32);
              mx = mx.max(x + j as i32);
          }
          ans
      }
  }
  
  

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