Propagación de Plagas en Matrices usando BFS en Java

Solucion en c++

#include <iostream>
#include <queue>
#include <vector>
#include <unordered_set>

using namespace std;

vector<pair<int, int>> foundRottingOranges(vector<vector<int>> &grid)
{
    vector<pair<int, int>> rottenOranges;
    for (int i = 0; i < grid.size(); i++)
    {
        for (int j = 0; j < grid[i].size(); j++)
        {
            if (grid[i][j] == 2)
            {
                rottenOranges.push_back({i, j});
            }
        }
    }
    return rottenOranges;
}

int rottingOranges(vector<vector<int>> &grid)
{
    vector<pair<int, int>> origins = foundRottingOranges(grid);

    if (origins.empty()) // No hay naranjas podridas
    {
        return -1;
    }
    queue<pair<int, int>> q;
    for (const auto &origin : origins)
    {
        q.push(origin);
    }

    unordered_set<int> visited;

    int days = 0;
    for (const auto &origin : origins)
    {
        visited.emplace(origin.first * grid[0].size() + origin.second);
    }

    while (!q.empty())
    {
        int size = q.size();
        for (int i = 0; i < size; i++)
        {
            pair<int, int> current = q.front();
            q.pop();

            int directions[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
            for (const auto &dir : directions)
            {
                int newRow = current.first + dir[0];
                int newCol = current.second + dir[1];
                int newPosition = newRow * grid[0].size() + newCol;

                if (newRow >= 0 && newRow < grid.size() && newCol >= 0 && newCol < grid[0].size() &&
                    grid[newRow][newCol] == 1 && visited.find(newPosition) == visited.end())
                {
                    q.push({newRow, newCol});
                    grid[newRow][newCol] = 2;
                    visited.insert(newPosition);
                }
            }
        }
        if (!q.empty())
        {
            days++;
        }
    }

    for (const auto &row : grid)
    {
        for (int cell : row)
        {
            if (cell == 1)
            {
                return -1;
            }
        }
    }

    return days;
}

int main()
{
    vector<vector<int>> grid{{2, 1, 1}
                            ,{1, 1, 0},
                             {0, 1, 2}};
    cout << rottingOranges(grid) << endl;
    return 0;
}

Algo que inclui, es un catch cuando buscas en un extremo, ya que en este caso te sales del array y una bandera para que sume por una vez en la cola

from queue import Queue

STEPS = [(0, 1), (0, -1), (1, 0), (-1, 0)]


def rotting_oranges(farm: list[list[int]]) -> int:
    if farm is None or len(farm) == 0:
        return 0
    bad_tomatoes = Queue()
    good_tomatoes = 0
    for i in range(len(farm)):
        for j in range(len(farm[0])):
            if farm[i][j] == 2:
                bad_tomatoes.put((i, j))
            if farm[i][j] == 1:
                good_tomatoes += 1
    if good_tomatoes == 0:
        return 0
    days = 1
    while not bad_tomatoes.empty():
        should_sum = False

        position = bad_tomatoes.get()
        for step in STEPS:
            next_position = (position[0] + step[0], position[1] + step[1])
            try:
                if farm[next_position[0]][next_position[1]] == 1:
                    should_sum = True
                    farm[next_position[0]][next_position[1]] = 2
                    good_tomatoes -= 1
                    bad_tomatoes.put(next_position)
            except IndexError:
                continue
        if should_sum:
            days += 1
    return -1 if good_tomatoes != 0 else days


if __name__ == '__main__':
    assert rotting_oranges([[2, 1, 1], [1, 1, 0], [0, 1, 1]]) == 4
    assert rotting_oranges([[1, 1, 1], [1, 1, 0], [0, 1, 1]]) == -1
    assert rotting_oranges([[0, 1, 1], [1, 1, 0], [0, 1, 2]]) == 5