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Steady-Gradient 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 trail is surveyed at evenly spaced markers, and heights[i] is the elevation at marker i.

A stretch is a contiguous run of markers heights[i] through heights[j]. A stretch is steady when it covers at least three markers and the elevation change between each pair of neighbouring markers inside it is the same. The shared change may be positive, negative or zero.

Return how many steady stretches the trail has. Two stretches are different when they start at different markers or end at different markers, even if their elevations match.

Examples

Example 1

Input: heights = [3, 6, 9, 12]
Output: 3

Every neighbouring pair rises by 3. The stretches covering at least three markers are markers 0 to 2, markers 1 to 3 and markers 0 to 3.

Example 2

Input: heights = [1, 2, 4, 8]
Output: 0

The three changes are 1, 2 and 4. No stretch of three or more markers has one repeated change.

Example 3

Input: heights = [5, 5, 5, 5]
Output: 3

The trail is flat, so every change is 0 and the stretches markers 0 to 2, 1 to 3 and 0 to 3 are all steady.

Constraints

1 <= heights.length <= 5000
-1000 <= heights[i] <= 1000
The answer is at most 1.3 * 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_steady_stretches(heights: list[int]) -> int:

Java

public int countSteadyStretches(int[] heights)

The intended approach

Time O(n) for n markers, a single pass.Space O(1).

Scan the markers from the third one onwards, keeping how many steady stretches end exactly at the previous marker. When the change into the current marker equals the change into the previous one, every one of those stretches can be extended by the current marker and one brand new three-marker stretch appears, so that count goes up by one and is added to the running total. Any other marker breaks the run and resets the count to zero.

Pythonentry point: count_steady_stretches

def count_steady_stretches(heights):
    total = 0
    ending_here = 0
    # a marker that keeps the change equal to the previous change extends every
    # steady stretch that ended at the previous marker, and opens one more
    for i in range(2, len(heights)):
        if heights[i] - heights[i - 1] == heights[i - 1] - heights[i - 2]:
            ending_here += 1
            total += ending_here
        else:
            ending_here = 0
    return total

The obvious version

Fix each starting marker, read off the change to the next marker, then walk forwards while the change holds, counting one stretch for every marker reached from the third onwards. Quadratic in the number of markers.

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_count_steady_stretches(heights):
    n = len(heights)
    total = 0
    for i in range(n - 2):
        step = heights[i + 1] - heights[i]
        for j in range(i + 2, n):
            if heights[j] - heights[j - 1] != step:
                break
            total += 1
    return total

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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