0007MediumArrayHash TablePrefix Sum

Trail Elevation Stretches

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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 has split a hiking trail into numbered segments laid end to end. deltas[i] is the signed elevation change in metres across segment i: positive on a climb, negative on a descent, zero across flat ground.

A stretch is one or more segments hiked back to back, with no gaps. The net elevation change of a stretch is the sum of the changes of its segments.

Given deltas and an integer target, return how many stretches have a net elevation change of exactly target. Two stretches count separately whenever they begin or end on different segments, even if their nets match and even if one sits inside the other.

A stretch always covers at least one segment; there is no empty stretch.

Examples

Example 1

Input: deltas = [40, -40, 40], target = 40
Output: 3

Segment 0 alone, segment 2 alone, and all three segments together each net 40 metres. The descent in the middle cancels the first climb, which is why the full trail also qualifies.

Example 2

Input: deltas = [12, 5, -5, 7], target = 12
Output: 2

Segment 0 on its own nets 12, and segments 0 through 2 net 12 as well because the 5 metre rise is undone by the 5 metre drop.

Example 3

Input: deltas = [-3, -3, -3], target = -6
Output: 2

Segments 0 and 1 form one qualifying stretch and segments 1 and 2 form another. They overlap, so both are counted.

Constraints

1 <= deltas.length <= 2 * 10^4
-1000 <= deltas[i] <= 1000
-10^7 <= target <= 10^7

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 count_elevation_stretches(deltas: list[int], target: int) -> int:

Java

public int countElevationStretches(int[] deltas, int target)

The intended approach

Time O(n)Space O(n)

Track the running elevation from the trailhead to the end of each segment. A stretch nets `target` exactly when the running elevation at its end minus the running elevation just before its start equals `target`, so as the sweep advances it asks how many earlier running elevations equal `current - target` and adds that many stretches. A tally keyed by running elevation supplies those counts in constant time, seeded with one occurrence of 0 for the trailhead so stretches starting at segment 0 are included. Because descents are allowed, running elevations can revisit the same value, and each revisit is why a tally of counts is needed rather than a single stored position.

Pythonentry point: count_elevation_stretches

def count_elevation_stretches(deltas, target):
    tally = {0: 1}
    running = 0
    stretches = 0
    for change in deltas:
        running += change
        stretches += tally.get(running - target, 0)
        tally[running] = tally.get(running, 0) + 1
    return stretches

The obvious version

Sum every contiguous stretch explicitly, 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_stretches(deltas, target):
    n = 0
    for i in range(len(deltas)):
        run = 0
        for j in range(i, len(deltas)):
            run += deltas[j]
            if run == target:
                n += 1
    return n

How this problem was checked

Reference solution executed against 12 test cases — 12 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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