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Sync Reachability Across Calibration Bands

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Trains the technique from

LeetCode 3532Path Existence Queries in a Graph I

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 rack holds n instruments numbered 0 to n - 1. Instrument i reports the calibration reading nums[i], and the readings are handed over already sorted in non-decreasing order.

Two instruments can talk to each other directly when their readings sit within maxDiff of one another, that is when abs(nums[i] - nums[j]) <= maxDiff. A calibration signal may also be relayed: it travels from one instrument to another whenever some chain of direct links connects them, with any number of hops in between.

Each entry of queries is a pair [u, v]. Return an array of booleans, one per query in the same order, saying whether a signal can get from instrument u to instrument v. An instrument can always reach itself.

Examples

Example 1

Input: n = 5, nums = [2, 3, 4, 6, 9], maxDiff = 2, queries = [[0, 3], [0, 4], [4, 4], [3, 4]]
Output: [true, false, true, false]

Readings 2, 3, 4 and 6 form one relay chain because each neighbouring step is 1, 1 and 2. The reading 9 is 3 above 6 and further from the rest, so nothing links it, and a query onto itself is still true.

Example 2

Input: n = 3, nums = [5, 5, 7], maxDiff = 0, queries = [[0, 1], [1, 2], [0, 2]]
Output: [true, false, false]

Only identical readings can talk when maxDiff is 0, so instruments 0 and 1 are linked and instrument 2 stands alone.

Example 3

Input: n = 3, nums = [0, 10, 20], maxDiff = 10, queries = [[0, 2], [0, 1]]
Output: [true, true]

Instrument 0 links to 1 and 1 links to 2, each step being exactly 10, so a relayed signal covers 0 to 2 even though their own readings differ by 20.

Example 4

Input: n = 6, nums = [0, 1, 7, 8, 9, 40], maxDiff = 3, queries = [[0, 1], [0, 2], [2, 4], [4, 5]]
Output: [true, false, true, false]

Instruments 0 and 1 sit together with a step of 1. Instruments 2, 3 and 4 form a second group, their neighbouring steps being 1 and 1. Instrument 5 reads 40, which is 31 above its nearest neighbour, so it links to nothing.

Constraints

1 <= n == nums.length <= 10^5
0 <= nums[i] <= 10^5
nums is sorted in non-decreasing order.
0 <= maxDiff <= 10^5
1 <= queries.length <= 10^5
queries[i] == [u_i, v_i]
0 <= u_i, v_i < n

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 sync_reachability(count: int, readings: list[int], max_gap: int, pairs: list[list[int]]) -> list[bool]:

Java

public boolean[] syncReachability(int count, int[] readings, int maxGap, int[][] pairs)

The intended approach

Time O(n + q)Space O(n) for the band labels plus O(q) for the returned list

Because the readings arrive sorted, a group of mutually relayable instruments is exactly a run of consecutive positions in which every neighbouring step is at most max_gap: any wider step splits the rack, and inside a run each instrument reaches the next one directly. One sweep stamps a band number on each position, and each query is then a comparison of two band numbers.

Pythonentry point: sync_reachability

def sync_reachability(count, readings, max_gap, pairs):
    # The readings arrive sorted, so a direct link can only ever exist between
    # neighbours or across a run of neighbours. One sweep labels each maximal run
    # whose consecutive steps are all within max_gap.
    # Time O(n + q). Space O(n) plus the O(q) answer.
    band = [0] * count
    label = 0
    for i in range(1, count):
        if readings[i] - readings[i - 1] > max_gap:
            label += 1
        band[i] = label
    return [band[u] == band[v] for u, v in pairs]

The obvious version

Build every direct link by comparing all pairs of readings, then run a breadth-first search per query.

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_sync_reachability(count, readings, max_gap, pairs):
    from collections import deque
    adj = [[] for _ in range(count)]
    for i in range(count):
        for j in range(i + 1, count):
            if abs(readings[i] - readings[j]) <= max_gap:
                adj[i].append(j)
                adj[j].append(i)
    out = []
    for u, v in pairs:
        seen = [False] * count
        seen[u] = True
        q = deque([u])
        while q:
            x = q.popleft()
            for y in adj[x]:
                if not seen[y]:
                    seen[y] = True
                    q.append(y)
        out.append(seen[v])
    return out

How this problem was checked

Reference solution executed against 7 test cases — 7 passed
4 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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