0043MediumArrayBinary Search

Survey Line Crest

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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 drone flies one straight line and stores the ground elevation under each of its sample points in elevations, in flight order. Elevations are metres relative to sea level, so they may be negative. No two neighbouring samples came out equal.

Call index i a crest when its elevation beats each neighbour it has: elevations[i] > elevations[i - 1] whenever i > 0, and elevations[i] > elevations[i + 1] whenever i + 1 < elevations.length. Read the ground beyond either end of the line as infinitely deep, which is why a crest always exists.

Return the index of a crest. Several samples may qualify and any one of their indices is accepted. Reading elevations off the drone is slow, so your routine must consult only O(log n) of the samples.

Examples

Example 1

Input: elevations = [-8, -3]
Output: 1

The last sample has one neighbour and sits above it, and the ground past the end counts as infinitely deep.

Example 2

Input: elevations = [12, 40, 15, 6]
Output: 1

Sample 1 rises above both 12 and 15, and no other index does that, so it is the only crest.

Example 3

Input: elevations = [1, 7, 2, 9, 5]
Output: 3

Index 3 beats 2 and 5. Index 1 also beats both of its neighbours, so returning 1 would be accepted as well.

Constraints

1 <= elevations.length <= 1000
-2^31 <= elevations[i] <= 2^31 - 1
elevations[i] != elevations[i + 1] for every valid i

More than one answer is valid — return any one of them.

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 crest_index(elevations: list[int]) -> int:

Java

public int crestIndex(int[] elevations)

The intended approach

Time O(log n)Space O(1)

Hold a window that is guaranteed to contain a crest, starting as the whole line. Compare the middle sample with the one just after it: if the ground is still rising there, the climb has to turn over somewhere to the right, so keep the right half; otherwise the middle sample is already at or past a turnover and the left half, middle included, still holds a crest. The window halves each round, so only O(log n) samples are read.

Pythonentry point: crest_index

def crest_index(elevations):
    lo, hi = 0, len(elevations) - 1
    while lo < hi:
        mid = (lo + hi) // 2
        if elevations[mid] < elevations[mid + 1]:
            lo = mid + 1
        else:
            hi = mid
    return lo

The obvious version

Read every sample and return the first index whose neighbours are both lower, treating a missing neighbour at either end as infinitely deep. O(n) reads, which the statement forbids, but it is plainly correct. It reports the leftmost crest, one of the accepted answers, and the fuzz generator emits lines with a single crest so the two routines must land on the very same index.

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_crest_index(elevations):
    n = len(elevations)
    for i in range(n):
        left_lower = i == 0 or elevations[i - 1] < elevations[i]
        right_lower = i == n - 1 or elevations[i + 1] < elevations[i]
        if left_lower and right_lower:
            return i
    return -1

How this problem was checked

Reference solution executed against 13 test cases — 13 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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