0278MediumArrayDynamic ProgrammingBreadth-First SearchMatrix

Steps to the Nearest Air Shaft

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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 mine level is surveyed as a rectangular grid of cells. grid[r][c] is 0 when the cell holds a ventilation shaft and 1 when it is plain rock. At least one cell holds a shaft.

A surveyor moves between cells that share an edge, one step per move, and every cell of the level can be walked through whatever its marking. Diagonal moves are not possible.

For every cell report the fewest steps from that cell to a cell holding a shaft; a cell that holds a shaft reports 0. Return these counts in a grid of the same shape as grid.

Examples

Example 1

Input: grid = [[0, 1, 1], [1, 1, 1], [1, 1, 1]]
Output: [[0, 1, 2], [1, 2, 3], [2, 3, 4]]

The one shaft sits at row 0, column 0. The cell at row 1, column 1 reaches it with one step up and one step left, so it reports 2, and the opposite corner reports 4.

Example 2

Input: grid = [[1, 1, 0], [1, 1, 1]]
Output: [[2, 1, 0], [3, 2, 1]]

The shaft sits at row 0, column 2. The cell at row 1, column 0 walks right, right and up, so it reports 3.

Example 3

Input: grid = [[0]]
Output: [[0]]

The single cell holds a shaft, so no steps are needed.

Constraints

m == grid.length
n == grid[i].length
1 <= m, n <= 10^4
1 <= m * n <= 10^4
grid[i][j] is either 0 or 1.
There is at least one 0 in grid.

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 steps_to_nearest_shaft(grid: list[list[int]]) -> list[list[int]]:

Java

public int[][] stepsToNearestShaft(int[][] grid)

The intended approach

Time O(m * n)Space O(m * n) for the returned grid and the queue

Seed a single queue with every shaft cell at count 0, then spread outward one ring at a time, writing a count into a cell the first time it is reached and queueing it. Because all the shafts entered the queue together, the first arrival at any cell is over a shortest route, so no cell is ever revisited or lowered later. The queue is a list with a moving head index rather than a recursive walk, and the answer grid doubles as the visited marker.

Pythonentry point: steps_to_nearest_shaft

def steps_to_nearest_shaft(grid):
    # Every shaft is seeded into the queue before the walk starts, so the wave
    # spreads from all of them at once and the first arrival at a cell is over a
    # shortest route.  A plain list with a moving head index serves as the queue,
    # so nothing recurses however large the level is.
    # Time: O(m * n), each cell is queued exactly once.
    # Space: O(m * n) for the answer grid and the queue.
    rows = len(grid)
    cols = len(grid[0])
    steps = [[-1] * cols for _ in range(rows)]
    queue = []
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == 0:
                steps[r][c] = 0
                queue.append((r, c))
    head = 0
    while head < len(queue):
        r, c = queue[head]
        head += 1
        nxt = steps[r][c] + 1
        for nr, nc in ((r - 1, c), (r + 1, c), (r, c - 1), (r, c + 1)):
            if 0 <= nr < rows and 0 <= nc < cols and steps[nr][nc] < 0:
                steps[nr][nc] = nxt
                queue.append((nr, nc))
    return steps

The obvious version

Collect the coordinates of every shaft, then for each cell take the smallest row difference plus column difference against all of them. Nothing blocks movement, so that difference is the step count for a straight staircase route.

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_steps_to_nearest_shaft(grid):
    rows = len(grid)
    cols = len(grid[0])
    shafts = [(r, c) for r in range(rows) for c in range(cols) if grid[r][c] == 0]
    out = []
    for r in range(rows):
        row = []
        for c in range(cols):
            best = None
            for sr, sc in shafts:
                d = abs(r - sr) + abs(c - sc)
                if best is None or d < best:
                    best = d
            row.append(best)
        out.append(row)
    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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