Skip to content

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

Solution 1: Enumeration

We can enumerate $j$ from left to right while maintaining the maximum value of $values[i] + i$ for elements to the left of $j$, denoted as $mx$. For each $j$, the maximum score is $mx + values[j] - j$. The answer is the maximum of these maximum scores for all positions.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{values}$. The space complexity is $O(1)$.

1
2
3
4
5
6
7
class Solution:
    def maxScoreSightseeingPair(self, values: List[int]) -> int:
        ans = mx = 0
        for j, x in enumerate(values):
            ans = max(ans, mx + x - j)
            mx = max(mx, x + j)
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution {
    public int maxScoreSightseeingPair(int[] values) {
        int ans = 0, mx = 0;
        for (int j = 0; j < values.length; ++j) {
            ans = Math.max(ans, mx + values[j] - j);
            mx = Math.max(mx, values[j] + j);
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution {
public:
    int maxScoreSightseeingPair(vector<int>& values) {
        int ans = 0, mx = 0;
        for (int j = 0; j < values.size(); ++j) {
            ans = max(ans, mx + values[j] - j);
            mx = max(mx, values[j] + j);
        }
        return ans;
    }
};

1
2
3
4
5
6
7
8
func maxScoreSightseeingPair(values []int) (ans int) {
    mx := 0
    for j, x := range values {
        ans = max(ans, mx+x-j)
        mx = max(mx, x+j)
    }
    return
}

1
2
3
4
5
6
7
8
function maxScoreSightseeingPair(values: number[]): number {
    let [ans, mx] = [0, 0];
    for (let j = 0; j < values.length; ++j) {
        ans = Math.max(ans, mx + values[j] - j);
        mx = Math.max(mx, values[j] + j);
    }
    return ans;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
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
    }
}

Comments