0913EasyArray

Peaks Behind a Tall Neighbour

Tracked in this browser only

Trains the technique from

LeetCode 3285Find Indices of Stable Mountains

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.

Peak heights along a ridge are given as height, and a cut-off is given as threshold.

A peak is sheltered when the peak directly before it on the ridge stands strictly higher than the cut-off. The first peak has nothing before it, so it is never sheltered.

Return the positions of the sheltered peaks, in increasing order.

Examples

Example 1

Input: height = [9, 3, 14, 6, 11], threshold = 7
Output: [1, 3]

Position 1 sits behind the peak of 9, which clears 7. Position 2 sits behind the peak of 3, which does not. Position 3 sits behind 14 and position 4 behind 6, so only the first of those counts.

Example 2

Input: height = [3, 3, 3, 3], threshold = 3
Output: []

Every peak matches the cut-off without clearing it, so nothing is sheltered.

Example 3

Input: height = [5, 6, 7, 8, 9], threshold = 4
Output: [1, 2, 3, 4]

Every peak clears the cut-off, so every position from 1 onwards is sheltered.

Constraints

2 <= height.length <= 100
1 <= height[i] <= 100
1 <= threshold <= 100

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 stable_mountains(height: list[int], threshold: int) -> list[int]:

Java

public List<Integer> stableMountains(int[] height, int threshold)

The intended approach

Time O(n) for n peaks, one pass.Space O(1) beyond the answer.

Whether a peak is sheltered depends on one other peak only, the one directly before it, so there is nothing to accumulate and nothing to look ahead for. Walk the ridge from the second peak onwards and record the position whenever the peak behind it clears the cut-off. Walking forwards means the positions come out in increasing order already.

Pythonentry point: stable_mountains

def stable_mountains(height, threshold):
    out = []
    for i in range(1, len(height)):
        if height[i - 1] > threshold:
            out.append(i)
    return out

The obvious version

Take every position in turn and compare it against every other position on the ridge, recording the position whenever the one it is paired with is both directly behind it and above the cut-off. Quadratic in the peak count, and it never relies on walking the ridge in order.

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_stable_mountains(height, threshold):
    n = len(height)
    out = []
    for i in range(n):
        for j in range(n):
            if j + 1 == i and height[j] > threshold:
                out.append(i)
    return out

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

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