0468HardArrayMathGeometrySliding WindowSorting

Aiming the Yard Camera

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Trains the technique from

LeetCode 1610Maximum Number of Visible Points

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 stocktake camera sits on a fixed mast in a container yard. The yard is mapped on integer grid coordinates: the mast stands at mast = [mx, my], and tags lists the coordinates of the radio tags stuck to the containers, so tags[i] = [xi, yi]. Several tags may sit at the same coordinates, and a tag may sit at the mast's own coordinates.

The camera can be swivelled to any bearing but its lens is fixed: it takes in a wedge that is exactly sweep degrees wide, with the mast at the wedge's tip. Aim it once and it reads every tag that falls inside that wedge, edges included; a tag exactly on either edge is read. Containers do not hide one another, so distance and stacking are irrelevant. A tag lying at the mast's own coordinates has no bearing at all and is read no matter where the camera points.

Return the largest number of tags a single aim can read.

Examples

Example 1

Input: tags = [[6, 4], [7, 4], [8, 5]], sweep = 45, mast = [5, 4]
Output: 3

Pointing the camera due east takes in all three tags at once, so nothing is left out of the reading.

Example 2

Input: tags = [[5, 4], [5, 4], [2, 8], [8, 8]], sweep = 0, mast = [5, 4]
Output: 3

Two tags share the mast's coordinates and are read on any aim. The wedge has no width at all, so of the two remaining tags only one can be lined up.

Example 3

Input: tags = [[44, 51], [44, 49], [46, 52]], sweep = 40, mast = [50, 50]
Output: 3

One aim towards the west, tilted slightly north, brings all three tags inside the wedge together.

Constraints

1 <= tags.length <= 10^5
tags[i].length == 2
mast.length == 2
0 <= sweep < 360
0 <= mx, my, xi, yi <= 100
The answer is a plain tag count, so no rounding is involved.

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 tags_in_view(tags: list[list[int]], sweep: int, mast: list[int]) -> int:

Java

public int tagsInView(int[][] tags, int sweep, int[] mast)

The intended approach

Time O(n log n), dominated by sorting the bearingsSpace O(n) for the bearing list and its doubled copy

Only the bearing of each tag from the mast matters, so turn every tag into its bearing in degrees and set aside the tags standing on the mast, which are always read. Sort the bearings and append a second copy with a full turn added, which lets a wedge that crosses the zero bearing appear as a plain contiguous run. Then slide a window over that doubled list, pulling the trailing end forward while the spread exceeds `sweep`, and keep the widest run seen. Add back the tags on the mast at the end.

Pythonentry point: tags_in_view

import math

EDGE = 1e-7


def tags_in_view(tags, sweep, mast):
    # Bearings of distinct lattice directions differ by at least 0.0029 degrees,
    # so EDGE only ever forgives a tag sitting exactly on a wedge edge.
    # time O(n log n) for the sort, space O(n) for the bearing list
    mx, my = mast[0], mast[1]
    on_mast = 0
    bearings = []
    for tag in tags:
        dx = tag[0] - mx
        dy = tag[1] - my
        if dx == 0 and dy == 0:
            on_mast += 1
        else:
            bearings.append(math.degrees(math.atan2(dy, dx)))
    bearings.sort()
    ring = bearings + [b + 360.0 for b in bearings]
    limit = sweep + EDGE
    best = 0
    lo = 0
    for hi in range(len(ring)):
        while ring[hi] - ring[lo] > limit:
            lo += 1
        if hi - lo + 1 > best:
            best = hi - lo + 1
    return best + on_mast

The obvious version

Take each tag's bearing in turn as the wedge's trailing edge, then count how many tags fall within `sweep` degrees of it going anticlockwise, measuring each gap with a modulo of a full turn. Keep the largest count and add the tags standing on the mast.

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

import math


def bf_tags_in_view(tags, sweep, mast):
    mx, my = mast[0], mast[1]
    on_mast = 0
    bearings = []
    for tag in tags:
        dx = tag[0] - mx
        dy = tag[1] - my
        if dx == 0 and dy == 0:
            on_mast += 1
        else:
            bearings.append(math.degrees(math.atan2(dy, dx)))
    best = 0
    for start in bearings:
        seen = 0
        for other in bearings:
            gap = (other - start) % 360.0
            if gap <= sweep + 1e-7:
                seen += 1
        if seen > best:
            best = seen
    return best + on_mast

How this problem was checked

Reference solution executed against 8 test cases — 8 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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