write a C++ program to 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 So we have to 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:

write a C++ program to 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
So we have to 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.
Input: arr[][C] = { {2, 1, 0, 2, 1},
Tahir Iqbal Department of Computer Sciences. BULC
{0, 0, 1, 2, 1},
{1, 0, 0, 2, 1}};
Output:
All oranges cannot be rotten.
Below is algorithm.
1) Create an empty Q.
2) Find all rotten oranges and enqueue them to Q. Also enqueue
a delimiter to indicate beginning of next time frame.
3) While Q is not empty do following
3.a) While delimiter in Q is not reached
(i) Dequeue an orange from queue, rot all adjacent oranges.
While rotting the adjacents, make sure that time frame
is incremented only once. And time frame is not icnremented
if there are no adjacent oranges.
3.b) Dequeue the old delimiter and enqueue a new delimiter. The
oranges rotten in previous time frame lie between the two
delimiters.