0487MediumArrayHash TableSimulation

Sweeper Drone Sortie

Tracked in this browser only

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 floor-sweeping drone sits on its charging pad at cell (0, 0) of a warehouse floor that stretches away without limit in every direction. It begins the sortie facing north, the direction in which y grows.

The drone works through moves in order. Each entry is one of the following:

-2: pivot a quarter turn to its left without leaving the cell it is on;
-1: pivot a quarter turn to its right without leaving the cell it is on;
a value k between 1 and 9: creep k cells straight ahead, one cell per step.

pillars[i] = [x, y] marks a cell taken up by a roof pillar. Before each single-cell step the drone looks at the cell directly ahead: if a pillar stands there it stays where it is and gives up the rest of that entry, but it still carries out every entry that follows.

Return the largest value of x * x + y * y over every cell the drone occupies during the sortie, where (x, y) is the cell.

Examples

Example 1

Input: moves = [2, -1, 6], pillars = [[0, 2]]
Output: 37

The drone steps north to (0, 1), finds the pillar ahead and gives up that entry, pivots right to face east, then creeps to (6, 1), where 6 * 6 + 1 * 1 = 37.

Example 2

Input: moves = [5, -2, -2, 3], pillars = []
Output: 25

The drone reaches (0, 5) with nothing in the way, then two left pivots turn it to face south and it comes back to (0, 2); the furthest cell it occupied was (0, 5).

Example 3

Input: moves = [-2, 5, -1, 3], pillars = [[-3, 0], [2, 2]]
Output: 13

Facing west the drone halts at (-2, 0) with the pillar at (-3, 0) ahead of it, then pivots back to north and finishes at (-2, 3), where 4 + 9 = 13.

Constraints

1 <= moves.length <= 10^4
moves[i] is -2, -1, or an integer in the range [1, 9].
0 <= pillars.length <= 10^4
pillars[i].length == 2
-3 * 10^4 <= pillars[i][0], pillars[i][1] <= 3 * 10^4
No pillar stands on the charging pad at (0, 0). The same cell may be listed more than once.
The value returned is at most 10^10.

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 sweep_reach(moves: list[int], pillars: list[list[int]]) -> int:

Java

public long sweepReach(int[] moves, int[][] pillars)

The intended approach

Time O(total steps + p)Space O(p)

Drop every pillar cell into a hash set once, then walk the entries. A pivot rotates the heading vector, mapping `(dx, dy)` to `(-dy, dx)` for a left turn and to `(dy, -dx)` for a right one. A creep entry is carried out one cell at a time, testing the cell ahead against the set and abandoning that entry on a hit. Compare the squared distance against the running best after each accepted step.

Pythonentry point: sweep_reach

def sweep_reach(moves, pillars):
    # Keep the pillar cells in a hash set so each single-cell step is a constant
    # time lookup, and track the best squared distance as the drone goes.
    # time O(sum of step counts + p), space O(p)
    blocked = set()
    for cell in pillars:
        blocked.add((cell[0], cell[1]))
    x = y = 0
    dx, dy = 0, 1
    best = 0
    for move in moves:
        if move == -2:
            dx, dy = -dy, dx
        elif move == -1:
            dx, dy = dy, -dx
        else:
            for _ in range(move):
                nx, ny = x + dx, y + dy
                if (nx, ny) in blocked:
                    break
                x, y = nx, ny
                d = x * x + y * y
                if d > best:
                    best = d
    return best

The obvious version

Hold the heading as an index into a four-entry table of unit offsets, and test each candidate cell by scanning the whole pillar list. Every cell the drone occupies is appended to a log, and the answer is the largest squared distance in that log.

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_sweep_reach(moves, pillars):
    headings = [(0, 1), (1, 0), (0, -1), (-1, 0)]
    facing = 0
    x = y = 0
    visited = [(0, 0)]
    for move in moves:
        if move == -2:
            facing = (facing - 1) % 4
        elif move == -1:
            facing = (facing + 1) % 4
        else:
            for _ in range(move):
                nx = x + headings[facing][0]
                ny = y + headings[facing][1]
                hit = False
                for cell in pillars:
                    if cell[0] == nx and cell[1] == ny:
                        hit = True
                        break
                if hit:
                    break
                x, y = nx, ny
                visited.append((x, y))
    return max(px * px + py * py for px, py in visited)

How this problem was checked

Reference solution executed against 11 test cases — 11 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in