0012MediumArrayBreadth-First SearchMatrix

Rack Corruption Sweep

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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 storage aisle is wired as a rectangular grid racks. Every cell carries one of three codes:

0 marks an empty bay,
1 marks a node running healthy firmware,
2 marks a node whose firmware is already corrupted.

Corruption travels on a fixed clock. During each tick, every healthy node that shares an edge with a node corrupted before that tick becomes corrupted itself. Empty bays carry nothing, and cells touching only at a corner are not neighbours.

Return how many ticks pass before nothing healthy is left in the aisle. If some healthy node can never be reached, return -1 instead.

Examples

Example 1

Input: racks = [[1, 1, 0], [1, 2, 1], [0, 1, 1]]
Output: 2

The four edge neighbours of the corrupted middle go in tick 1, and the two remaining corner nodes go in tick 2.

Example 2

Input: racks = [[2, 0, 1]]
Output: -1

The empty bay sits between the two nodes, so the healthy one on the right is never touched.

Example 3

Input: racks = [[0, 0], [2, 0]]
Output: 0

No healthy node is present when the clock starts, so the aisle is already settled.

Constraints

rows == racks.length
cols == racks[i].length
1 <= rows, cols <= 10
Each racks[i][j] is 0, 1, or 2

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 infection_ticks(racks: list[list[int]]) -> int:

Java

public int infectionTicks(int[][] racks)

The intended approach

Time O(rows * cols)Space O(rows * cols)

Every already-corrupted node starts spreading at the same moment, so seed a breadth first search with all of them at once and advance one whole layer per tick. Track how many healthy nodes remain; when the layers run dry with a positive count left, those nodes are cut off by empty bays and the answer is -1.

Pythonentry point: infection_ticks

def infection_ticks(racks):
    rows, cols = len(racks), len(racks[0])
    frontier = []
    healthy = 0
    for r in range(rows):
        for c in range(cols):
            if racks[r][c] == 2:
                frontier.append((r, c))
            elif racks[r][c] == 1:
                healthy += 1
    ticks = 0
    while frontier and healthy:
        nxt = []
        for r, c in frontier:
            for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                nr, nc = r + dr, c + dc
                if 0 <= nr < rows and 0 <= nc < cols and racks[nr][nc] == 1:
                    racks[nr][nc] = 2
                    healthy -= 1
                    nxt.append((nr, nc))
        frontier = nxt
        ticks += 1
    return -1 if healthy else ticks

The obvious version

Simulate on a copy by rescanning the entire grid each tick for healthy cells that touch a corrupted one, flipping them together, and stopping when a pass changes nothing. O((rows * cols)^2).

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_infection_ticks(racks):
    grid = [row[:] for row in racks]
    rows, cols = len(grid), len(grid[0])
    ticks = 0
    while True:
        doomed = []
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] != 1:
                    continue
                for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                    nr, nc = r + dr, c + dc
                    if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] == 2:
                        doomed.append((r, c))
                        break
        if not doomed:
            break
        for r, c in doomed:
            grid[r][c] = 2
        ticks += 1
    for row in grid:
        if 1 in row:
            return -1
    return ticks

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

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