0248EasyArraySorting

Tightest Gauge Gaps

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Trains the technique from

LeetCode 1200Minimum Absolute Difference

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 calibration bench measures a batch of gauges against a reference block and records readings, the signed offset in micrometres that each gauge showed. The readings come in the order the gauges were measured, which is not necessarily sorted, and no two gauges showed the same offset.

Call the tightest gap the smallest difference between two of the recorded offsets.

Return every pair of offsets whose difference equals the tightest gap. Write each pair as [lower, higher] and list the pairs by their lower offset, smallest first. No two reported pairs share a lower offset, so that order is unambiguous.

Examples

Example 1

Input: readings = [3, 8, 1, 4, 9]
Output: [[3, 4], [8, 9]]

The tightest gap is 1 micrometre, and two pairs of offsets sit that close: 3 with 4, and 8 with 9. The pair with the lower first offset is listed first.

Example 2

Input: readings = [-7, -2, 0, -9]
Output: [[-9, -7], [-2, 0]]

The tightest gap is 2 micrometres, reached by -9 with -7 and by -2 with 0.

Example 3

Input: readings = [5, -5]
Output: [[-5, 5]]

Two gauges leave a single pair, whose difference of 10 micrometres is therefore the tightest gap.

Constraints

2 <= readings.length <= 10^5
-10^6 <= readings[i] <= 10^6
All offsets in readings are distinct

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 tightest_gaps(readings: list[int]) -> list[list[int]]:

Java

public List<List<Integer>> tightestGaps(int[] readings)

The intended approach

Time O(n log n) for n readings, dominated by the sort.Space O(n) for the sorted copy, plus the returned pairs.

Sort the offsets, which puts the two members of any close pair next to each other. Sweep the sorted offsets once to learn the tightest neighbouring gap, then sweep again and keep every neighbouring pair that matches it; that second sweep runs left to right, so the pairs come out already ordered by their lower offset.

Pythonentry point: tightest_gaps

def tightest_gaps(readings):
    ordered = sorted(readings)
    tightest = ordered[1] - ordered[0]
    for i in range(2, len(ordered)):
        gap = ordered[i] - ordered[i - 1]
        if gap < tightest:
            tightest = gap
    pairs = []
    for i in range(1, len(ordered)):
        if ordered[i] - ordered[i - 1] == tightest:
            pairs.append([ordered[i - 1], ordered[i]])
    return pairs

The obvious version

Measure every unordered pair of offsets to find the smallest difference, then walk every pair again to collect the ones that match and sort the collected pairs. 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_tightest_gaps(readings):
    n = len(readings)
    tightest = None
    for i in range(n):
        for j in range(i + 1, n):
            gap = abs(readings[i] - readings[j])
            if tightest is None or gap < tightest:
                tightest = gap
    hits = []
    for i in range(n):
        for j in range(i + 1, n):
            low = min(readings[i], readings[j])
            high = max(readings[i], readings[j])
            if high - low == tightest:
                hits.append([low, high])
    hits.sort()
    return hits

How this problem was checked

Reference solution executed against 10 test cases — 10 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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