0367MediumArrayTwo PointersBinary SearchSorting

Smallest Mast Reach

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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 pipeline runs dead straight. sensors gives the positions of the sensor nodes bolted along it and masts gives the positions of the relay masts. Positions are whole numbers of metres, they arrive in no particular order, and two entries may share a position.

Every mast is configured with the same reach r. A mast at position p then covers the stretch from p - r to p + r, endpoints included, and a sensor counts as covered when at least one mast reaches it.

Return the smallest reach r that leaves every sensor covered.

Examples

Example 1

Input: sensors = [7,2,9], masts = [5]
Output: 4

The single mast sits at 5, so the three sensors are 2, 3 and 4 metres away. A reach of 4 covers all three.

Example 2

Input: sensors = [12,3,8], masts = [10,1]
Output: 2

The sensor at 3 is 2 metres from the mast at 1, the sensor at 8 is 2 metres from the mast at 10, and the sensor at 12 is also 2 metres from the mast at 10.

Example 3

Input: sensors = [9], masts = [2,100]
Output: 7

The sensor at 9 is 7 metres from the mast at 2 and 91 metres from the mast at 100, so a reach of 7 covers it.

Example 4

Input: sensors = [5,5,5], masts = [5,9]
Output: 0

Every sensor shares a position with a mast, so no reach at all is needed.

Constraints

1 <= sensors.length, masts.length <= 3 * 10^4
1 <= sensors[i], masts[i] <= 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 find_radius(sensors: list[int], masts: list[int]) -> int:

Java

public int findRadius(int[] sensors, int[] masts)

The intended approach

Time O(m log m + n log m) where n is the sensor count and m the mast countSpace O(m) for the sorted copy of the masts

The reach that has to be configured is set by the sensor that is worst served, so for every sensor find the distance to its nearest mast and keep the largest of those distances. Sorting the masts once turns each of those lookups into a binary search for the insertion point, and the nearest mast is then whichever of the two masts either side of that point is closer.

Pythonentry point: find_radius

from bisect import bisect_left


def find_radius(sensors, masts):
    # Sorting the masts once lets every sensor be answered by a binary search.
    # Time O(m log m + n log m), space O(m) for the sorted copy of the masts.
    ordered = sorted(masts)
    worst = 0
    for position in sensors:
        k = bisect_left(ordered, position)
        best = None
        if k < len(ordered):
            best = ordered[k] - position
        if k > 0:
            left = position - ordered[k - 1]
            if best is None or left < best:
                best = left
        if best > worst:
            worst = best
    return worst

The obvious version

For each sensor scan every mast to find its nearest one, then report the largest of those per-sensor distances.

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_find_radius(sensors, masts):
    worst = 0
    for position in sensors:
        nearest = min(abs(position - m) for m in masts)
        if nearest > worst:
            worst = nearest
    return worst

How this problem was checked

Reference solution executed against 12 test cases — 12 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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