0794MediumArrayHash TableCounting

Isolated Dock Numbers

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Trains the technique from

LeetCode 2150Find All Lonely Numbers in the Array

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 bike-share audit lists the dock every bike was returned to yesterday. The list docks holds one dock number per return, so a dock number appears once for each bike returned to it.

Docks are numbered so that dock d - 1 and dock d + 1 are the two docks standing either side of dock d. Call a dock number isolated when it appears exactly once in docks and neither d - 1 nor d + 1 appears in docks at all.

Return every isolated dock number, in ascending order. Return an empty list when there is none.

Examples

Example 1

Input: docks = [42, 7, 19, 20]
Output: [7, 42]

Dock 7 was used once and docks 6 and 8 are absent, and dock 42 was used once with 41 and 43 absent. Docks 19 and 20 stand beside each other and both appear, so neither qualifies.

Example 2

Input: docks = [5, 6, 10, 10, 3]
Output: [3]

Dock 3 was used once with 2 and 4 absent. Dock 10 was used twice, and docks 5 and 6 sit next to each other.

Example 3

Input: docks = [42, 42]
Output: []

Dock 42 appears twice, so it is not isolated and nothing is reported.

Constraints

1 <= docks.length <= 10^5
0 <= docks[i] <= 10^6

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 isolated_docks(docks: list[int]) -> list[int]:

Java

public List<Integer> isolatedDocks(int[] docks)

The intended approach

Time O(n log n) for n returns: the tally is built in linear time and the qualifying numbers are sorted.Space O(n) for the tally plus the returned numbers.

Build one tally of dock number to number of returns. A dock qualifies when its own tally is exactly one and neither neighbouring number is a key of the tally, both of which are constant-time lookups. Collect the qualifying numbers and sort them so the answer comes back ascending.

Pythonentry point: isolated_docks

def isolated_docks(docks):
    # one tally answers both questions: how often a dock appears, and whether
    # either neighbouring dock appears at all
    seen = {}
    for d in docks:
        seen[d] = seen.get(d, 0) + 1
    out = []
    for d, times in seen.items():
        if times == 1 and (d - 1) not in seen and (d + 1) not in seen:
            out.append(d)
    out.sort()
    return out

The obvious version

For every entry in the log, walk the whole log again counting how many returns went to that dock and to each of the two docks beside it, then keep the entry when the three counts are one, zero and zero. O(n^2).

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_isolated_docks(docks):
    out = []
    for i, d in enumerate(docks):
        here = 0
        below = 0
        above = 0
        for other in docks:
            if other == d:
                here += 1
            elif other == d - 1:
                below += 1
            elif other == d + 1:
                above += 1
        if here == 1 and below == 0 and above == 0:
            out.append(d)
    out.sort()
    return out

How this problem was checked

Reference solution executed against 14 test cases — 14 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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