# 1021. Best Sightseeing Pair

Given an array `A`

of positive integers, `A[i]`

represents the value of the `i`

-th sightseeing spot, and two sightseeing spots `i`

and `j`

have distance `j - i`

between them.

The *score* of a pair (`i < j`

) of sightseeing spots is (`A[i] + A[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: **

```
[8,1,5,2,6]
```

**Output: **

11Explanation:i = 0, j = 2,`A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11`

**Note:**

`2 <= A.length <= 50000`

`1 <= A[i] <= 1000`

#### Thoughts:

Scan from left to right, record:

- best score so far()
- the best left element which could generate the largest score with the next element.
- As we move on to the next, gap ++, impact/score — with the prev. best left element

class Solution { /** * @param Integer[] $A * @return Integer */ function maxScoreSightseeingPair($A) { $bestResult = 0; // Best Score (A[i] + A[j] + i - j) $bestAI = 0; // The best choice of A[i] so far foreach($A as $a){ // Moved to next, gap++, then the "sightseeing value" -- $bestAI --; // Best result = $bestResult = max($bestResult, $bestAI + $a ); $bestAI = max($a, $bestAI); } return $bestResult; } }

well sure I can combine $bestAI –; into other expressions. Just to make it clear in this way.

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