0731HardArrayTwo PointersBinary SearchDynamic ProgrammingGreedyBit ManipulationGraph TheorySorting

Relay Hops Across The Field Network

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Trains the technique from

LeetCode 3534Path Existence Queries in a Graph II

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 survey team has planted n radio relays, labelled 0 through n - 1. Relay i is tuned to the frequency level[i].

Two relays can pass a message straight to each other exactly when the absolute difference between their frequencies is at most maxGap. That is the only thing that decides it: labels, distance on the ground and the order the relays were planted play no part, and every pair of relays whose frequencies are close enough can pass messages both ways.

Each query queries[t] = [a, b] asks for the smallest number of direct passes a message needs to travel from relay a to relay b. If no chain of passes joins them, the answer for that query is -1. When a and b are the same relay the answer is 0.

Return an array holding the answer to every query, in the order the queries are listed.

Examples

Example 1

Input: n = 3, level = [0, 1, 10], maxGap = 5, queries = [[0, 2], [0, 1], [2, 2]]
Output: [-1, 1, 0]

Relays 0 and 1 are 1 apart, so they pass messages straight to each other. Relay 2 is 9 above relay 1 and 10 above relay 0, both beyond the gap, so nothing joins it to the rest. The last query starts and ends at the same relay.

Example 2

Input: n = 5, level = [0, 2, 4, 100, 102], maxGap = 2, queries = [[0, 2], [0, 4], [4, 3], [2, 0]]
Output: [2, -1, 1, 2]

Relay 0 at 0 and relay 2 at 4 are 4 apart, too far for one pass, but relay 1 at 2 is within 2 of both, so a message can go 0, 1, 2. Relay 4 at 102 and relay 3 at 100 are 2 apart, so they pass directly, and no relay bridges the gap from 4 up to 100.

Example 3

Input: n = 4, level = [7, 7, 7, 7], maxGap = 0, queries = [[0, 3], [1, 1]]
Output: [1, 0]

All four relays share a frequency, so any two of them differ by 0, which a gap of 0 allows. A query between two different relays takes one pass, and one that starts where it ends takes none.

Constraints

1 <= level.length <= 10^5
n == level.length
0 <= level[i] <= 10^5
0 <= maxGap <= 10^5
1 <= queries.length <= 10^5
queries[i].length == 2
0 <= queries[i][j] <= 10^5 - 1
Every label listed in queries is the label of a relay in the network.

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 relay_hop_counts(n: int, level: list[int], maxGap: int, queries: list[list[int]]) -> list[int]:

Java

public int[] relayHopCounts(int n, int[] level, int maxGap, int[][] queries)

The intended approach

Time O(n log n + q log n) for n relays and q queries.Space O(n log n) for the doubling table, plus O(q) for the answers.

Sort the relays by frequency. A step between neighbours in that order larger than `maxGap` cuts the network in two, so record which run of the sorted order each relay falls in and answer `-1` whenever the two ends of a query fall in different runs. Inside one run, a message that has to climb the order is best served by hopping to the furthest relay still within `maxGap`, which a two-pointer sweep records as `reach[p]`. Powers-of-two compositions of `reach` then let each query count its hops by taking the largest jumps that do not overshoot the target and adding one final hop, which is logarithmic per query. Frequencies decide the links symmetrically, so a query whose start sits above its target is answered by swapping the two ends.

Pythonentry point: relay_hop_counts

def relay_hop_counts(n, level, maxGap, queries):
    # Order the relays by frequency.  A link exists exactly between two relays whose
    # frequencies are within maxGap, so in this order the network splits into runs
    # separated by a step larger than maxGap, and two relays talk only inside one
    # run.  Inside a run, reaching a relay further up the order is done by always
    # hopping to the furthest relay still in range: reach[p] is that target, and a
    # doubling table over reach counts the hops from one position to another in
    # logarithmic time.  A frequency below the other end is handled by ordering the
    # two ends of the query, since the two directions need the same hop count.
    # Time O(n log n + q log n).  Space O(n log n).
    order = sorted(range(n), key=lambda i: level[i])
    rank = [0] * n
    for p, relay in enumerate(order):
        rank[relay] = p
    vals = [level[relay] for relay in order]

    run = [0] * n
    for p in range(1, n):
        run[p] = run[p - 1] + (1 if vals[p] - vals[p - 1] > maxGap else 0)

    reach = [0] * n
    j = 0
    for p in range(n):
        if j < p:
            j = p
        while j + 1 < n and vals[j + 1] - vals[p] <= maxGap:
            j += 1
        reach[p] = j

    depth = max(1, n.bit_length())
    jump = [reach]
    for _ in range(1, depth):
        prev = jump[-1]
        jump.append([prev[prev[p]] for p in range(n)])

    answers = []
    for pair in queries:
        low = rank[pair[0]]
        high = rank[pair[1]]
        if low > high:
            low, high = high, low
        if run[low] != run[high]:
            answers.append(-1)
            continue
        hops = 0
        cur = low
        for k in range(depth - 1, -1, -1):
            if jump[k][cur] < high:
                cur = jump[k][cur]
                hops += 1 << k
        if cur != high:
            hops += 1
        answers.append(hops)
    return answers

The obvious version

Compare every pair of relays to build the neighbour lists outright, then answer each query with its own breadth-first search from the starting relay, reporting the recorded distance of the target or -1 when the search never reached it.

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_relay_hop_counts(n, level, maxGap, queries):
    from collections import deque
    neighbours = [[] for _ in range(n)]
    for i in range(n):
        for j in range(i + 1, n):
            if abs(level[i] - level[j]) <= maxGap:
                neighbours[i].append(j)
                neighbours[j].append(i)
    out = []
    for pair in queries:
        start = pair[0]
        goal = pair[1]
        seen = [-1] * n
        seen[start] = 0
        q = deque([start])
        while q:
            cur = q.popleft()
            if cur == goal:
                break
            for nxt in neighbours[cur]:
                if seen[nxt] == -1:
                    seen[nxt] = seen[cur] + 1
                    q.append(nxt)
        out.append(seen[goal])
    return out

How this problem was checked

Reference solution executed against 12 test cases — 12 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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