0201MediumArrayBreadth-First SearchMatrix

Fewest Tiles Across the Floor

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Trains the technique from

LeetCode 1091Shortest Path in Binary Matrix

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 stockroom floor is an n x n grid of square tiles given as floor. floor[r][c] is 0 when the tile is free and 1 when a pallet stands on it, blocking the tile.

A picking robot has to get from the tile at the top-left corner to the tile at the bottom-right corner. From its current tile the robot can roll onto any of the up to eight tiles that touch it, including the four that only touch at a corner, as long as that tile is free. A route is the sequence of tiles the robot stands on, starting at the top-left tile and ending at the bottom-right tile, and every tile on it must be free.

Return the number of tiles on the shortest such route, counting both the tile it starts on and the tile it ends on. If no route exists, return -1.

Examples

Example 1

Input: floor = [[0, 0, 1], [1, 0, 1], [1, 0, 0]]
Output: 3

The robot rolls corner to corner through the middle tile: (0,0), then (1,1), then (2,2). Three tiles is the fewest possible on any board wider than one tile.

Example 2

Input: floor = [[0, 1, 1, 1], [0, 0, 1, 1], [1, 0, 0, 1], [1, 1, 0, 0]]
Output: 4

Pallets pen the robot into a staircase of free tiles, and the corner steps (0,0), (1,1), (2,2), (3,3) walk straight down it.

Example 3

Input: floor = [[0, 1, 0], [1, 1, 0], [0, 0, 0]]
Output: -1

All three tiles touching the starting corner carry pallets, so the robot cannot leave it and no route exists.

Constraints

n == floor.length
n == floor[i].length
1 <= n <= 100
floor[i][j] is 0 or 1

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 fewest_tiles_across(floor: list[list[int]]) -> int:

Java

public int fewestTilesAcross(int[][] floor)

The intended approach

Time O(n^2)Space O(n^2)

Every roll costs the same, so the tiles fan out from the start in rings of equal route length and a first-in-first-out sweep visits them in order of that length. Push the starting tile with a count of one, then repeatedly take the oldest tile off the front and push each free untouched neighbour with one more tile counted, marking it the moment it is queued so no tile is ever queued twice. The first time the bottom-right tile comes off the front its count is the shortest route, and an exhausted queue means it is walled off.

Pythonentry point: fewest_tiles_across

from collections import deque


def fewest_tiles_across(floor):
    # Time: O(n^2) -- each tile is queued at most once and has 8 neighbours.
    # Space: O(n^2) for the visited grid and the queue.
    n = len(floor)
    if floor[0][0] == 1 or floor[n - 1][n - 1] == 1:
        return -1
    steps = [(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]
    reached = [[False] * n for _ in range(n)]
    reached[0][0] = True
    queue = deque([(0, 0, 1)])
    while queue:
        r, c, tiles = queue.popleft()
        if r == n - 1 and c == n - 1:
            return tiles
        for dr, dc in steps:
            nr = r + dr
            nc = c + dc
            if 0 <= nr < n and 0 <= nc < n and not reached[nr][nc] and floor[nr][nc] == 0:
                reached[nr][nc] = True
                queue.append((nr, nc, tiles + 1))
    return -1

The obvious version

Hold a best-known tile count for every tile and keep sweeping the whole grid, lowering any tile that a neighbour can beat, until a full sweep changes nothing. No queue, no visit order to reason about.

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_fewest_tiles_across(floor):
    n = len(floor)
    if floor[0][0] == 1 or floor[n - 1][n - 1] == 1:
        return -1
    huge = n * n + 5
    count = [[huge] * n for _ in range(n)]
    count[0][0] = 1
    moving = True
    while moving:
        moving = False
        for r in range(n):
            for c in range(n):
                if floor[r][c] == 1:
                    continue
                for dr in (-1, 0, 1):
                    for dc in (-1, 0, 1):
                        if dr == 0 and dc == 0:
                            continue
                        nr = r + dr
                        nc = c + dc
                        if 0 <= nr < n and 0 <= nc < n and floor[nr][nc] == 0:
                            if count[nr][nc] + 1 < count[r][c]:
                                count[r][c] = count[nr][nc] + 1
                                moving = True
    best = count[n - 1][n - 1]
    return -1 if best >= huge else best

How this problem was checked

Reference solution executed against 16 test cases — 16 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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