0178MediumArrayStackMonotonic Stack

Loop Trail Next Rise

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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.

A survey crew walked a closed loop trail and wrote down the signed elevation change at each marker, in walking order, as the array deltas. The trail closes on itself, so once you step past the final marker you are back at marker 0.

Stand at a marker and keep walking forward around the loop until you reach a marker whose elevation change is strictly larger than the one you started from. Report that larger change. If you make a full circuit without finding one, report -1 instead.

Return an array answer of the same length, where answer[i] is what you report from marker i. Elevation changes may themselves be negative, and -1 is a perfectly legal change, so a reported -1 carries no extra meaning: it may be a change of -1 that you walked to, or it may be the signal that the loop holds nothing larger.

Examples

Example 1

Input: deltas = [2, 7, 1]
Output: [7, -1, 2]

Marker 0 finds 7 one step ahead. Marker 1 holds the largest change on the loop, so it reports -1. Marker 2 has to wrap past the end to reach 2.

Example 2

Input: deltas = [-4, -9, -2]
Output: [-2, -2, -1]

Both of the first two markers walk forward to -2. Marker 2 already holds the largest change, so it reports -1.

Example 3

Input: deltas = [6, 6, 6]
Output: [-1, -1, -1]

An equal change is not a rise, so no marker ever finds one.

Constraints

1 <= deltas.length <= 10^4
-10^9 <= deltas[i] <= 10^9

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 next_rise(deltas: list[int]) -> list[int]:

Java

public int[] nextRise(int[] deltas)

The intended approach

Time O(n)Space O(n)

Walk two laps of the trail while holding a stack of markers that have not yet found a rise, kept in decreasing order of change. Each new marker settles every stack entry it beats. Only the first lap pushes, so the second lap exists purely to serve the markers that need to wrap around.

Pythonentry point: next_rise

def next_rise(deltas):
    # One monotonic-stack pass over two laps of the loop.
    # Time O(n): every index is pushed once and popped at most once.
    # Space O(n) for the stack of indices still waiting for a rise.
    n = len(deltas)
    answer = [-1] * n
    waiting = []
    for step in range(2 * n):
        i = step % n
        while waiting and deltas[waiting[-1]] < deltas[i]:
            answer[waiting.pop()] = deltas[i]
        if step < n:
            waiting.append(i)
    return answer

The obvious version

For each marker, walk forward one step at a time modulo n until a strictly larger change turns up or the circuit closes. O(n^2).

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_next_rise(deltas):
    n = len(deltas)
    out = []
    for i in range(n):
        found = -1
        for step in range(1, n):
            j = (i + step) % n
            if deltas[j] > deltas[i]:
                found = deltas[j]
                break
        out.append(found)
    return out

How this problem was checked

Reference solution executed against 11 test cases — 11 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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