0590MediumArrayDynamic Programming

Best Pair of Lookout Boards

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Interpretive boards stand one step apart along a ridge trail, numbered 0 upward. Board i carries a review score values[i], which may be negative when visitors disliked it.

A guide picks an ordered pair of boards: an earlier board i and a later board j, with i strictly before j. The pair is rated by adding the two review scores together and then taking away the number of steps walked between the two boards, which is j - i.

Return the highest rating any such pair reaches.

Examples

Example 1

Input: values = [5, 4, 4, 4, 4, 100]
Output: 103

Pairing board 4 with board 5 adds 4 and 100 and takes away the single step between them, giving 103. No other pair reaches that.

Example 2

Input: values = [-3, -4]
Output: -8

Only one pair exists. Adding -3 and -4 gives -7, and one step is walked, so the rating is -8.

Example 3

Input: values = [1, -500, 1000]
Output: 999

Boards 0 and 2 add to 1001 and are two steps apart, for 999. Boards 1 and 2 add to 500 and are one step apart, for 499, and boards 0 and 1 rate -500.

Constraints

2 <= values.length <= 5 * 10^4
-1000 <= values[i] <= 1000

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def max_score_sightseeing_pair(values: list[int]) -> int:

Java

public int maxScoreSightseeingPair(int[] values)

The intended approach

Time O(n) — a single pass.Space O(1) — two scalars.

Split the rating so each board contributes on its own: the earlier board contributes its score plus its number, the later board its score minus its number. So walk the trail once, and at each board treat it as the later half of a pair by adding it to the best earlier contribution recorded so far. Then record this board's own earlier contribution for the boards ahead. The running best starts at the first board so that every pair considered has a genuine earlier partner, which matters because scores can be negative.

Pythonentry point: max_score_sightseeing_pair

def max_score_sightseeing_pair(values):
    # O(n) time, O(1) space: for a fixed later board j the score splits into a
    # part that depends only on j and a part that depends only on the earlier
    # board, so one pass keeps the best earlier part seen so far and combines it
    # with each j in turn.
    best_earlier = values[0] + 0
    answer = None
    for j in range(1, len(values)):
        score = best_earlier + values[j] - j
        if answer is None or score > answer:
            answer = score
        candidate = values[j] + j
        if candidate > best_earlier:
            best_earlier = candidate
    return answer

The obvious version

Rate every ordered pair of boards directly with two nested loops and keep the largest rating.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_max_score_sightseeing_pair(values):
    n = len(values)
    best = None
    for i in range(n):
        for j in range(i + 1, n):
            score = values[i] + values[j] - (j - i)
            if best is None or score > best:
                best = score
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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