0766MediumArrayMathGeometry

Sensors Within Each Scan Radius

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Trains the technique from

LeetCode 1828Queries on Number of Points Inside a Circle

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 grid carries fixed sensors. sensors[i] = [x, y] gives the integer coordinates of sensor i; two sensors may share a position.

A radar makes a series of circular sweeps. scans[j] = [x, y, r] describes a sweep centred at (x, y) with radius r. A sensor is covered by that sweep when its straight-line distance to the centre is at most r. A sensor lying exactly on the circle of radius r counts as covered.

Go through the sweeps in the order listed, work out how many sensors each one covers, and give back those totals as a single list. Two sensors on the same position are tallied separately.

Examples

Example 1

Input: sensors = [[7, 2], [2, 9], [10, 10], [6, 6]], scans = [[6, 3, 2], [9, 9, 3], [0, 0, 4]]
Output: [1, 1, 0]

The first sweep reaches sensor (7, 2), which sits a distance of about 1.41 from (6, 3). The second reaches sensor (10, 10), about 1.41 from (9, 9). The third covers no sensor at all, since the nearest is (6, 6) at about 8.49 from the origin.

Example 2

Input: sensors = [[3, 4], [0, 5], [4, 4]], scans = [[0, 0, 5]]
Output: [2]

Sensors (3, 4) and (0, 5) are both exactly 5 from the centre, so both count as covered. Sensor (4, 4) is about 5.66 away and is not.

Example 3

Input: sensors = [[2, 2], [2, 2], [7, 7]], scans = [[2, 2, 1], [7, 7, 9]]
Output: [2, 3]

The first sweep is centred on the position two sensors share, so it covers both of them. The second sweep has a radius of 9 and reaches all three sensors.

Constraints

1 <= sensors.length <= 500
sensors[i].length == 2
0 <= sensors[i][j] <= 500
1 <= scans.length <= 500
scans[j].length == 3
0 <= scans[i][j] <= 500
Sweep j is [x_j, y_j, r_j] with 1 <= r_j <= 500.
All coordinates and radii are integers.

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 sensors_in_range(sensors: list[list[int]], scans: list[list[int]]) -> list[int]:

Java

public int[] sensorsInRange(int[][] sensors, int[][] scans)

The intended approach

Time O(q * p) for q sweeps and p sensors.Space O(q) for the returned counts.

Handle each sweep on its own and test every sensor against it. Comparing the squared distance `dx*dx + dy*dy` with `r*r` keeps the whole test in integers, so no rounding can push a sensor on the circle to the wrong side, and `<=` is what makes the edge count as covered.

Pythonentry point: sensors_in_range

def sensors_in_range(sensors, scans):
    # Compare squared distances so the whole check stays in integers.
    # Time O(q * p) for q scans and p sensors, space O(q) for the answer.
    out = []
    for cx, cy, r in scans:
        reach = r * r
        total = 0
        for sx, sy in sensors:
            dx = sx - cx
            dy = sy - cy
            if dx * dx + dy * dy <= reach:
                total += 1
        out.append(total)
    return out

The obvious version

Index the sensor positions in a table that remembers how many sensors share each cell, then for each sweep walk every integer cell of the square that encloses the circle, keep the cells the circle actually covers, and add up the sensors sitting on them. Costs work proportional to the area of each sweep.

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_sensors_in_range(sensors, scans):
    tally = {}
    for x, y in sensors:
        key = (x, y)
        tally[key] = tally.get(key, 0) + 1
    out = []
    for cx, cy, r in scans:
        reach = r * r
        total = 0
        for x in range(cx - r, cx + r + 1):
            dx = x - cx
            room = reach - dx * dx
            if room < 0:
                continue
            for y in range(cy - r, cy + r + 1):
                dy = y - cy
                if dy * dy <= room:
                    total += tally.get((x, y), 0)
        out.append(total)
    return out

How this problem was checked

Reference solution executed against 9 test cases — 9 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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