The problem you're describing is a classic example of a breadth-first search (BFS) problem in graph theory. Here's a step-by-step solution:

1. First, we need to identify the initial infected persons (cells with value 2) and add them to a queue. We also need to count the number of virus-free persons (cells with value 1).

2. Then, we start a loop that continues until the queue is empty. In each iteration of the loop, we remove a cell from the queue and check its four adjacent cells. If an adjacent cell is virus-free, we infect it (change its value to 2), decrement the count of virus-free persons, and add it to the queue.

3. We also keep track of the number of days passed. Since the virus spreads to all adjacent virus-free persons in one day, the number of days is equal to the number of iterations of the loop.

4. The loop ends when either the queue is empty or there are no more virus-free persons. If there are no more virus-free persons, the number of days is the minimum number of days required for everyone to be infected. If the queue is empty but there are still virus-free persons, it means that these persons are isolated and cannot be infected, so we return -1.

Here's a Python code snippet that implements this solution:

```python
from collections import deque

def min_days(grid):
rows, cols = len(grid), len(grid[0])
queue = deque()
free_count = 0

for r in range(rows):
for c in range(cols):
if grid[r][c] == 2:
queue.append((r, c, 0))
elif grid[r][c] == 1:
free_count += 1

if free_count == 0:
return 0

directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
days = 0

while queue:
r, c, days = queue.popleft()

for dr, dc in directions:
nr, nc = r + dr, c + dc

if 0

Question

The problem you're describing is a classic example of a breadth-first search (BFS) problem in graph theory. Here's a step-by-step solution:

1. First, we need to identify the initial infected persons (cells with value 2) and add them to a queue. We also need to count the number of virus-free persons (cells with value 1).

2. Then, we start a loop that continues until the queue is empty. In each iteration of the loop, we remove a cell from the queue and check its four adjacent cells. If an adjacent cell is virus-free, we infect it (change its value to 2), decrement the count of virus-free persons, and add it to the queue.

3. We also keep track of the number of days passed. Since the virus spreads to all adjacent virus-free persons in one day, the number of days is equal to the number of iterations of the loop.

4. The loop ends when either the queue is empty or there are no more virus-free persons. If there are no more virus-free persons, the number of days is the minimum number of days required for everyone to be infected. If the queue is empty but there are still virus-free persons, it means that these persons are isolated and cannot be infected, so we return -1.

Here's a Python code snippet that implements this solution:

```python
from collections import deque

def min_days(grid):
    rows, cols = len(grid), len(grid[0])
    queue = deque()
    free_count = 0

for r in range(rows):
        for c in range(cols):
            if grid[r][c] == 2:
                queue.append((r, c, 0))
            elif grid[r][c] == 1:
                free_count += 1

if free_count == 0:
        return 0

directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
    days = 0

while queue:
        r, c, days = queue.popleft()

for dr, dc in directions:
            nr, nc = r + dr, c + dc

if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] == 1:
                grid[nr][nc] = 2
                free_count -= 1
                queue.append((nr, nc, days + 1))

if free_count == 0:
            return days

return -1
```

This function takes a 2D list `grid` as input and returns the minimum number of days required for everyone to be infected. If it's impossible to infect everyone, it returns -1.

Knowee AI · Accepted Answer