# 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

• 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;
}