0860EasyArrayEnumeration

Ridge Points Along the Profile

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 survey profile is given as mountain, where mountain[i] is the height at station i.

A station is a ridge point when it is strictly higher than the station on each side of it. The first and last stations are never ridge points, since each has only one neighbour.

Return the ridge point stations in increasing order.

Examples

Example 1

Input: mountain = [2, 5, 3, 6, 4]
Output: [1, 3]

Station 1 stands at 5 above the 2 and the 3 beside it, and station 3 stands at 6 above the 3 and the 4 beside it.

Example 2

Input: mountain = [1, 3, 3, 1]
Output: []

Stations 1 and 2 are both at height 3, so neither stands strictly above both of its neighbours.

Example 3

Input: mountain = [1, 2, 3]
Output: []

The only station with a neighbour on each side is station 1, and it stands below station 2.

Constraints

3 <= mountain.length <= 100
1 <= mountain[i] <= 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 ridge_points(mountain: list[int]) -> list[int]:

Java

public List<Integer> ridgePoints(int[] mountain)

The intended approach

Time O(n) for n stations: constant work each.Space O(1) beyond the returned list.

Being a ridge point depends only on a station and its two neighbours, so one walk over the stations that have both is enough. Check each against the height on either side and collect the ones that stand above both, which comes out in increasing order for free.

Pythonentry point: ridge_points

def ridge_points(mountain):
    out = []
    for i in range(1, len(mountain) - 1):
        if mountain[i] > mountain[i - 1] and mountain[i] > mountain[i + 1]:
            out.append(i)
    return out

The obvious version

For every station, gather the heights of whichever neighbours it has, and keep the station when it has two of them and stands above the larger. Builds the neighbour list explicitly rather than restricting the walk in advance.

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_ridge_points(mountain):
    n = len(mountain)
    out = []
    for i in range(n):
        around = []
        if i - 1 >= 0:
            around.append(mountain[i - 1])
        if i + 1 < n:
            around.append(mountain[i + 1])
        if len(around) == 2 and mountain[i] > max(around):
            out.append(i)
    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

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