0115EasyArrayHash TableSorting

Core Sample Rank Labels

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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 drills a line of core samples and logs the elevation of each sample's bedrock contact in metres against the site datum, so a logged value may be below the datum, on it, or above it. The log arrives as readings in drilling order. A shift that drilled nothing at all leaves the log empty.

The written report hides the raw elevations and prints rank labels instead. Labels are assigned like this:

The deepest elevation on the log is labelled 1.
Samples logging the identical elevation carry the identical label.
Labels run without gaps: the next elevation upwards from a labelled one takes the next whole number.

Return the labels in drilling order, one per reading.

Examples

Example 1

Input: readings = [-15, 40, -15, 8]
Output: [1, 3, 1, 2]

Three distinct elevations appear, -15 then 8 then 40, so they take labels 1, 2 and 3, and the repeated -15 keeps label 1 in both positions.

Example 2

Input: readings = [0, -4, -4, 7, 7, 12]
Output: [2, 1, 1, 3, 3, 4]

The distinct elevations run -4, 0, 7, 12, so 0 lands on label 2 while each repeated elevation reuses its own label.

Example 3

Input: readings = [-7, -8]
Output: [2, 1]

Only two samples were drilled and -8 is the deeper of the pair, so it takes label 1 and -7 takes label 2.

Constraints

0 <= readings.length <= 10^5
-10^9 <= readings[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 rank_core_samples(readings: list[int]) -> list[int]:

Java

public int[] rankCoreSamples(int[] readings)

The intended approach

Time O(n log n)Space O(n)

Collapse the log to its distinct elevations and sort them, which puts them in label order by construction. Store each distinct elevation against its position in that order, then read the log again and swap in the stored label for each reading.

Pythonentry point: rank_core_samples

def rank_core_samples(readings):
    ladder = sorted(set(readings))
    place = {}
    for i, value in enumerate(ladder):
        place[value] = i + 1
    return [place[value] for value in readings]

The obvious version

For each reading, scan the whole log and collect the distinct elevations strictly below it; the label is one more than how many there are. O(n^3) in the worst case.

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_rank_core_samples(readings):
    out = []
    for value in readings:
        lower = []
        for other in readings:
            if other < value and other not in lower:
                lower.append(other)
        out.append(len(lower) + 1)
    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

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