0699EasyArrayHash TableCounting

Counting Label Pairs at a Fixed Gap

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Trains the technique from

LeetCode 2006Count Number of Pairs With Absolute Difference K

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 shelf of parts is described by labels, where labels[i] is the number printed on the part in slot i. Two different slots i and j form a matched pair when the two printed numbers differ by exactly gap, that is when abs(labels[i] - labels[j]) == gap.

Return how many matched pairs of slots the shelf holds. A pair of slots counts once, regardless of which slot you name first, and two slots printed with the same number are still two different slots.

Examples

Example 1

Input: labels = [3, 3, 3, 6, 6], gap = 3
Output: 6

Each of the three slots printed `3` pairs with each of the two slots printed `6`, since those numbers differ by `3`. Two slots printed with the same number differ by `0`, so they never pair here.

Example 2

Input: labels = [4, 9, 4, 9, 4], gap = 5
Output: 6

Three slots carry `4` and two carry `9`, and `9 - 4` is `5`, so every one of those slot combinations is a matched pair.

Example 3

Input: labels = [10, 20, 30, 40], gap = 10
Output: 3

The matched pairs are the slots printed `10` and `20`, `20` and `30`, and `30` and `40`. Slots printed `10` and `30` differ by `20`, so they do not count.

Constraints

1 <= labels.length <= 200
1 <= labels[i] <= 100
1 <= gap <= 99

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 count_gap_pairs(labels: list[int], gap: int) -> int:

Java

public int countGapPairs(int[] labels, int gap)

The intended approach

Time O(n) for n slots, one pass with constant-time tally lookups.Space O(n) for the tally.

Sweep the slots left to right keeping a tally of how many times each number has already been seen. When the current number is `v`, the slots it pairs with are exactly the earlier ones printed `v - gap` or `v + gap`, so add those two tallies to the running total and then record `v`. Each pair is closed once, by its later slot.

Pythonentry point: count_gap_pairs

def count_gap_pairs(labels, gap):
    tally = {}
    total = 0
    # walking left to right, every earlier label that sits `gap` away from the
    # current one closes exactly one pair
    for v in labels:
        total += tally.get(v - gap, 0) + tally.get(v + gap, 0)
        tally[v] = tally.get(v, 0) + 1
    return total

The obvious version

Test every pair of slots with two nested loops over the indices and count the ones whose printed numbers differ by exactly the gap.

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_count_gap_pairs(labels, gap):
    total = 0
    n = len(labels)
    for i in range(n):
        for j in range(i + 1, n):
            if abs(labels[i] - labels[j]) == gap:
                total += 1
    return total

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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