Minimum time required to rot 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

Determine what is the minimum time required so that all the oranges become rotten. A rotten orange at index [i,j] can rot other fresh orange at indexes [i-1,j], [i+1,j], [i,j-1], [i,j+1] (up, down, left and right). If it is impossible to rot every orange then simply return -1.

Examples: 

Input:  arr[][C] = { {2, 1, 0, 2, 1},
                     {1, 0, 1, 2, 1},
                     {1, 0, 0, 2, 1}};
Output:
All oranges can become rotten in 2-time frames.
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:
All oranges cannot be rotten.
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 Solution:  

• Approach: The idea is very basic. Traverse through all oranges in multiple rounds. In every round, rot the oranges to the adjacent position of oranges which were rotten in the last round.
• Algorithm: 
  1. Create a variable no = 2 and changed = false
  2. Run a loop until there is no cell of the matrix which is changed in an iteration.
  3. 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.
  4. Traverse the matrix and check if there is any cell which is 1. If 1 is present return -1
  5. Else return no – 2
• Implementation:

Max time incurred: 2

• Complexity Analysis: 
  • 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)).
  • Space Complexity:O(1). 
    No extra space is required.

Efficient Solution 

• Approach: 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 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 that time can be reduced by using this efficient approach of BFS. 
   
• Algorithm: 
  1. Create an empty queue Q. 
     1. Find all rotten oranges and enqueue them to Q. Also, enqueue a delimiter to indicate the beginning of the next time frame.
     2. Run a loop While Q is not empty
     3. Do following while delimiter in Q is not reached
  2. Dequeue an orange from the queue, 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.
  3. Dequeue the old delimiter and enqueue a new delimiter. The oranges rotten in the previous time frame lie between the two delimiters.
• Dry run of the above approach:

• Implementation:

Time required for all oranges to rot => 2

• Complexity Analysis: 
  • 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), i.e. the number of elements. So time complexity is O(R *C).
  • Space Complexity: O(R*C). 
    To store the elements in a queue O(R*C) space is needed.

Thanks to Gaurav Ahirwar for suggesting the above solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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