1171MediumArrayDynamic ProgrammingQueueHeap (Priority Queue)Monotonic Queue

Hopping Along the Row of Pads

Tracked in this browser only

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.

A row of pads holds the scores scores. You begin on the first pad and must finish on the last one. From a pad you may hop forward to any pad at most k places ahead, so long as it is still on the row.

Your total is the sum of the scores of every pad you stand on, counting the first and the last.

Return the largest total possible.

Examples

Example 1

Input: scores = [1, -100, 2], k = 2
Output: 3

A hop of two places clears the heavy loss in the middle, so only the two ends are stood on.

Example 2

Input: scores = [1, -100, 2], k = 1
Output: -97

A hop of one place forces every pad to be stood on, the loss included.

Example 3

Input: scores = [1, 2, 3, 4, 5], k = 2
Output: 15

Every pad scores something, so standing on all of them is best, and hops of one place manage exactly that.

Constraints

1 <= scores.length <= 10^5
1 <= k <= 10^5
-10^4 <= scores[i] <= 10^4

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_result(scores: list[int], k: int) -> int:

Java

public int maxResult(int[] scores, int k)

The intended approach

Time O(n) over the row.Space O(n) for the totals and the queue.

The best total ending on a pad is its own score plus the best total ending on some pad within `k` places behind it. Working that out by looking at all `k` of them costs the row times the reach, which is too much when both run to a hundred thousand. The saving is that the window of candidate pads only ever slides forward, so the best inside it can be carried along. Keep a queue of pad numbers whose totals fall from front to back. The front is always the best in reach, so retire it once it drops out of reach behind. Before adding a new pad, drop every pad at the back whose total is no better, since a later pad with a total at least as good makes them useless forever. Each pad joins and leaves the queue once, so the whole row is one pass. Retiring the out-of-reach front matters: without it a pad too far back could still be handed on.

Pythonentry point: max_result

def max_result(scores, k):
    from collections import deque
    n = len(scores)
    best = [0] * n
    best[0] = scores[0]
    window = deque([0])
    for i in range(1, n):
        while window and window[0] < i - k:
            window.popleft()
        best[i] = best[window[0]] + scores[i]
        while window and best[window[len(window) - 1]] <= best[i]:
            window.pop()
        window.append(i)
    return best[n - 1]

The obvious version

Work along the row keeping the best total ending on each pad, and for every pad look back over all the pads within reach to find the best to hop from. Costs the row length times the reach.

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_result(scores, k):
    n = len(scores)
    big = -(1 << 60)
    best = [big] * n
    best[0] = scores[0]
    for i in range(1, n):
        lo = i - k
        if lo < 0:
            lo = 0
        top = big
        for j in range(lo, i):
            if best[j] > top:
                top = best[j]
        best[i] = top + scores[i]
    return best[n - 1]

How this problem was checked

Reference solution executed against 15 test cases — 15 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in