0123EasyArrayHash TableSorting

Repeated Elevation Marker

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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 stamps every benchmark it visits with that benchmark's height above the datum, in whole centimetres. Benchmarks sunk below the datum get a negative stamp, and one right on it gets 0. The stamps come back as the array marks, in visiting order.

A stamp is supposed to identify a benchmark on its own, so the crew needs to know whether any height was stamped on more than one benchmark. Return true if some value occurs at two or more positions of marks, and false if every stamp is distinct.

Heights the same distance either side of the datum are different heights: -40 and 40 are two separate stamps, not a repeat.

Examples

Example 1

Input: marks = [14, -3, 9, -3]
Output: true

The stamp -3 was used at the second benchmark and again at the last one, so the crew has a collision.

Example 2

Input: marks = [8, -20, 0, 5]
Output: false

Four different heights, one of them exactly at the datum, and no value comes back twice.

Example 3

Input: marks = [-6, 6]
Output: false

The two benchmarks sit the same distance from the datum but on opposite sides, so their stamps differ.

Constraints

1 <= marks.length <= 10^5
-10^9 <= marks[i] <= 10^9

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 repeated_elevation(marks: list[int]) -> bool:

Java

public boolean repeatedElevation(int[] marks)

The intended approach

Time O(n) expectedSpace O(n)

Sweep the stamps once, keeping every height already seen in a hash set. If the current height is already in the set a repeat exists and the sweep can stop immediately; otherwise record it and continue. Reaching the end without a hit means all stamps are distinct.

Pythonentry point: repeated_elevation

def repeated_elevation(marks):
    seen = set()
    for height in marks:
        if height in seen:
            return True
        seen.add(height)
    return False

The obvious version

Compare every pair of positions directly with nested loops and report the first match. O(n^2), no auxiliary structure.

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_repeated_elevation(marks):
    n = len(marks)
    for i in range(n):
        for j in range(i + 1, n):
            if marks[i] == marks[j]:
                return True
    return False

How this problem was checked

Reference solution executed against 9 test cases — 9 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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