# Rotting Oranges

#### You have been given a grid containing some oranges. Each cell of this grid has one of the three integers values:

#### Every second, any fresh orange that is adjacent(4-directionally) to a rotten orange becomes rotten.

#### Your task is to find out the minimum time after which no cell has a fresh orange. If it's impossible to rot all the fresh oranges then print -1.

##### Note:

```
1. The grid has 0-based indexing.
2. A rotten orange can affect the adjacent oranges 4 directionally i.e. Up, Down, Left, Right.
```

##### Input Format:

```
The first line of input contains two single space-separated integers 'N' and 'M' representing the number of rows and columns of the grid respectively.
The next 'N' lines contain 'M' single space-separated integers each representing the rows of the grid.
```

##### Output Format:

```
The only line of output contains a single integer i.e. The minimum time after which no cell has a fresh orange.
If it's impossible to rot all oranges, print -1.
```

##### Note:

```
You are not required to print the expected output, it has already been taken care of. Just implement the function.
```

##### Constraints:

```
1 <= N <= 500
1 <= M <= 500
0 <= grid[i][j] <= 2
Time Limit: 1 sec
```

The idea is very simple and naive. We will process the rotten oranges second by second. Each second, we rot all the fresh oranges that are adjacent to the already rotten oranges. The time by which there are no rotten oranges left to process will be our minimum time.

In the first traversal of the grid, we will process all the cells with value 2 (rotten oranges). We will also mark their adjacent cells as rotten for the next traversal. Now, we can’t mark them by assigning the same value i.e. 2 because then we won’t able to differentiate between the current processing cells and the cells which are going to be processed in the next traversal. So, we will mark them as value 3. More formally, we will be marking the adjacent cells as ‘CURR_ROTTEN’ + 1 in each traversal of the grid.

Here is the complete algorithm.

- Initialize ‘TIME’ to 0 and ‘CURR_ROTTEN’ to 2.
- Loop until there is no rotten orange left to process.
- Initialize ‘NOT_FOUND’ to true.
- We will run a nested loop and traverse the grid.
- If the element of the grid is equal to ‘CURR_ROTTEN’ then we have to process this rotten orange.
- So we will assign the adjacent elements to (‘CURR_ROTTEN’ + 1), if the adjacent orange is fresh (value is 1), and ‘NOT_FOUND’ to false.

- If ‘NOT_FOUND’ is true, break.
- Else, increment ‘TIME’ by 1.
- Increment ‘CURR_ROTTEN’ by 1.

- At last, we will traverse the grid and check if there is any fresh orange left i.e. a cell with value 1. If found, return -1.
- Else return time elapsed i.e. maximum of ‘TIME’ - 1 and 0.

We can use Breadth-First-Search which is similar to the level order traversal of a Binary Tree, to solve this problem.

We will be using a Queue data structure and inserting cells into this Queue level by level. Here, all the already rotten oranges will be at level 0, all the fresh oranges adjacent to them will be at level 1, all the fresh oranges adjacent to level 1 oranges will be at level 2, and so on. The time to reach the last level will be our minimum time.

Here, is the complete algorithm-

- Initialize ‘TIME’ to 0.
- Declare a Queue data structure and a 2D boolean array ‘VISITED’.
- Push all cells with rotten orange to the queue.
- Loop till queue is not empty
- Get the size of the current level/queue.
- Loop till the 'LEVEL_SIZE' is zero:
- Get the front cell of the queue.
- Push all adjacent cells with fresh oranges to the queue.
- Mark the cells visited which are pushed into the queue.

- Increment ‘TIME’ by 1.

- Iterate the grid again. If a fresh orange is still present, return -1.
- Return maximum of ‘TIME’ - 1 and 0.