Minimum time required to rotten all oranges

Given a matrix of dimension M * N, where each cell in the matrix can have values 0, 1 or 2 which has the following meaning:  

• 0: Empty cell
• 1: Cells have fresh oranges
• 2: Cells have rotten oranges

The task is to the minimum time required so that all the oranges become rotten. A rotten orange at index (i,j ) can rot other fresh oranges which are its neighbors (up, down, left, and right). If it is impossible to rot every orange then simply return -1.

Minimum time required to rotten all oranges

Examples: 

Input:  arr[][C] = { {2, 1, 0, 2, 1}, {1, 0, 1, 2, 1}, {1, 0, 0, 2, 1}};
Output: 2
Explanation: At 0th time frame:
{2, 1, 0, 2, 1}
{1, 0, 1, 2, 1}
{1, 0, 0, 2, 1}
At 1st time frame:
{2, 2, 0, 2, 2}
{2, 0, 2, 2, 2}
{1, 0, 0, 2, 2}
At 2nd time frame:
{2, 2, 0, 2, 2}
{2, 0, 2, 2, 2}
{2, 0, 0, 2, 2}

Input:  arr[][C] = { {2, 1, 0, 2, 1}, {0, 0, 1, 2, 1}, {1, 0, 0, 2, 1}}
Output: -1
Explanation: At 0th time frame:
{2, 1, 0, 2, 1}
{0, 0, 1, 2, 1}
{1, 0, 0, 2, 1}
At 1st time frame:
{2, 2, 0, 2, 2}
{0, 0, 2, 2, 2}
{1, 0, 0, 2, 2}
At 2nd time frame:
{2, 2, 0, 2, 2}
{0, 0, 2, 2, 2}
{1, 0, 0, 2, 2}
The 1 at the bottom left corner of the matrix is never rotten.

Naive Approach:  

Traverse through all oranges in multiple rounds. In every round, rot the oranges to the adjacent position of oranges that were rotten in the last round.

Follow the steps below to solve the problem:

• Create a variable no = 2 and changed = false.
• Run a loop until there is no cell of the matrix which is changed in an iteration.
• Run a nested loop and traverse the matrix: 
  • If the element of the matrix is equal to no then assign the adjacent elements to no + 1 if the adjacent element’s value is equal to 1, i.e. not rotten, and update changed to true.
• Traverse the matrix and check if there is any cell that is 1. 
  • If 1 is present return -1
• Else return no – 2.

Below is the implementation of the above approach.

Max time incurred: 2

Time Complexity: O((R*C) * (R *C)), 

• The matrix needs to be traversed again and again until there is no change in the matrix, that can happen max(R *C)/2 times. 
• So time complexity is O((R * C) * (R *C)).

Auxiliary Space: O(1), No extra space is required.

Minimum time required to rot all oranges using Breadth First Search:

The idea is to use Breadth First Search. The condition of oranges getting rotten is when they come in contact with other rotten oranges. This is similar to a breadth-first search where the graph is divided into layers or circles and the search is done from lower or closer layers to deeper or higher layers. 

In the previous approach, the idea was based on BFS but the implementation was poor and inefficient. To find the elements whose values are no the whole matrix had to be traversed. So time can be reduced by using this efficient approach of BFS.  

Follow the steps below to solve the problem:

• Create an empty queue Q. 
• Find all rotten oranges and enqueue them to Q. Also, enqueue a delimiter to indicate the beginning of the next time frame.
• Run a loop While Q is not empty and do the following while the delimiter in Q is not reached
  • Dequeue an orange from the queue, and rot all adjacent oranges. 
  • While rotting the adjacent, make sure that the time frame is incremented only once. And the time frame is not incremented if there are no adjacent oranges.
  • Dequeue the old delimiter and enqueue a new delimiter. The oranges rotten in the previous time frame lie between the two delimiters.
• Return the last time frame.

Illustration:

Step 0: At time t=0, insert all the rotten oranges into the queue, these will act as source for multisource BFS.

Minimum time required to rotten all oranges

Step 1: At time t=1, all the fresh oranges which are adjacent to rotten oranges all made rotten and inserted into the queue.

Minimum time required to rotten all oranges

Step 2: At time t=2, the last fresh orange is made rotten and inserted into the queue.

Minimum time required to rotten all oranges

Step 3: After the last fresh orange is made rotten, no new orange will be added to queue and queue will become empty. Since the last fresh orange became rotten at time t=2, minimum time required to rot all oranges will be 2.

Minimum time required to rotten all oranges

Below is the implementation of the above approach.

Time required for all oranges to rot => 1

Time Complexity: O( R *C), Each element of the matrix can be inserted into the queue only once so the upper bound of iteration is O(R*C)
Auxiliary Space: O(R*C), To store the elements in a queue.

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