0687HardArraySliding WindowSortingBucket SortOrdered Set

Close Samples Within a Short Window

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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 tilt meter writes one signed sample per tick, given as samples, where samples[i] is the reading at tick i. An alarm fires when two different ticks i and j satisfy both of these:

the ticks are close in time: abs(i - j) <= indexSpan;
the readings are close in value: abs(samples[i] - samples[j]) <= valueSpan.

Both limits are inclusive, so a gap of exactly indexSpan ticks counts, and a value difference of exactly valueSpan counts. Readings may be negative, and valueSpan may be 0, in which case only two equal readings can fire the alarm.

Return true if such a pair of ticks exists, and false otherwise.

Examples

Example 1

Input: samples = [10, 100, 11], indexSpan = 2, valueSpan = 1
Output: true

Ticks 0 and 2 are 2 apart, which the time limit allows, and their readings 10 and 11 differ by 1, which the value limit allows.

Example 2

Input: samples = [10, 100, 11], indexSpan = 1, valueSpan = 1
Output: false

With the time limit at 1 tick, only the pairs (0, 1) and (1, 2) may be considered, and those readings differ by 90 and 89.

Example 3

Input: samples = [-7, -4], indexSpan = 1, valueSpan = 3
Output: true

The two ticks are 1 apart and their readings differ by exactly 3, which the inclusive value limit allows.

Constraints

2 <= samples.length <= 10^5
-10^9 <= samples[i] <= 10^9
1 <= indexSpan <= 10^5
indexSpan is at most samples.length.
0 <= valueSpan <= 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 has_close_pair(samples: list[int], indexSpan: int, valueSpan: int) -> bool:

Java

public boolean hasClosePair(int[] samples, int indexSpan, int valueSpan)

The intended approach

Time O(n) for n ticks: constant work per tick.Space O(min(n, indexSpan)) for the live buckets.

Map each reading to the bucket `reading // (valueSpan + 1)`, using floor division so that negative readings are handled the same way as positive ones. Two readings in one bucket are automatically within `valueSpan`, and two readings within `valueSpan` must be in the same bucket or in neighbouring ones, so each new reading only has to be checked against the bucket it lands in and the two either side. Keeping only the readings of the last `indexSpan + 1` ticks in the table, and dropping the reading that just fell out of the window, enforces the time limit.

Pythonentry point: has_close_pair

def has_close_pair(samples, indexSpan, valueSpan):
    width = valueSpan + 1
    # one slot per bucket: two samples that land in the same bucket are already
    # within valueSpan of each other, so the answer is settled the moment that
    # happens and no bucket ever needs to hold two live samples
    live = {}
    for i, v in enumerate(samples):
        if i > indexSpan:
            gone = samples[i - indexSpan - 1]
            key = gone // width
            if live.get(key) == gone:
                del live[key]
        b = v // width          # floor division, so negatives bucket correctly
        if b in live:
            return True
        if b - 1 in live and v - live[b - 1] <= valueSpan:
            return True
        if b + 1 in live and live[b + 1] - v <= valueSpan:
            return True
        live[b] = v
    return False

The obvious version

For every tick, compare its reading against each of the following ticks that lie inside the time limit, and report the first pair that is also close in value. Quadratic 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_has_close_pair(samples, indexSpan, valueSpan):
    n = len(samples)
    for i in range(n):
        for j in range(i + 1, min(n, i + indexSpan + 1)):
            if abs(samples[i] - samples[j]) <= valueSpan:
                return True
    return False

How this problem was checked

Reference solution executed against 16 test cases — 16 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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