# 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```

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},
{0, 0, 1, 2, 1},
{1, 0, 0, 2, 1}};
Output:
All oranges cannot be rotten.

```

The idea is to user Breadth First Search. 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.```

Below is implementation of the above idea.

`// C++ program to find minimum time required to make all`
`// oranges rotten`
`#include<bits/stdc++.h>`
`#define R 3`
`#define C 5`
`using` `namespace` `std;`
`// function to check whether a cell is valid / invalid`
`bool` `isvalid(``int` `i, ``int` `j)`
`{`
`    ``return` `(i >= 0 && j >= 0 && i < R && j < C);`
`}`
`// structure for storing coordinates of the cell`
`struct` `ele {`
`    ``int` `x, y;`
`};`
`// Function to check whether the cell is delimiter`
`// which is (-1, -1)`
`bool` `isdelim(ele temp)`
`{`
`    ``return` `(temp.x == -1 && temp.y == -1);`
`}`
`// Function to check whether there is still a fresh`
`// orange remaining`
`bool` `checkall(``int` `arr[][C])`
`{`
`    ``for` `(``int` `i=0; i<R; i++)`
`       ``for` `(``int` `j=0; j<C; j++)`
`          ``if` `(arr[i][j] == 1)`
`             ``return` `true``;`
`    ``return` `false``;`
`}`
`// This function finds if it is possible to rot all oranges or not.`
`// If possible, then it returns minimum time required to rot all,`
`// otherwise returns -1`
`int` `rotOranges(``int` `arr[][C])`
`{`
`    ``// Create a queue of cells`
`    ``queue<ele> Q;`
`    ``ele temp;`
`    ``int` `ans = 0;`
`    ``// Store all the cells having rotten orange in first time frame`
`    ``for` `(``int` `i=0; i<R; i++)`
`    ``{`
`       ``for` `(``int` `j=0; j<C; j++)`
`       ``{`
`            ``if` `(arr[i][j] == 2)`
`            ``{`
`                ``temp.x = i;`
`                ``temp.y = j;`
`                ``Q.push(temp);`
`            ``}`
`        ``}`
`    ``}`
`    ``// Separate these rotten oranges from the oranges which will rotten`
`    ``// due the oranges in first time frame using delimiter which is (-1, -1)`
`    ``temp.x = -1;`
`    ``temp.y = -1;`
`    ``Q.push(temp);`
`    ``// Process the grid while there are rotten oranges in the Queue`
`    ``while` `(!Q.empty())`
`    ``{`
`        ``// This flag is used to determine whether even a single fresh`
`        ``// orange gets rotten due to rotten oranges in current time`
`        ``// frame so we can increase the count of the required time.`
`        ``bool` `flag = ``false``;`
`        ``// Process all the rotten oranges in current time frame.`
`        ``while` `(!isdelim(Q.front()))`
`        ``{`
`            ``temp = Q.front();`
`            ``// Check right adjacent cell that if it can be rotten`
`            ``if` `(isvalid(temp.x+1, temp.y) && arr[temp.x+1][temp.y] == 1)`
`            ``{`
`                ``// if this is the first orange to get rotten, increase`
`                ``// count and set the flag.`
`                ``if` `(!flag) ans++, flag = ``true``;`
`                ``// Make the orange rotten`
`                ``arr[temp.x+1][temp.y] = 2;`
`                ``// push the adjacent orange to Queue`
`                ``temp.x++;`
`                ``Q.push(temp);`
`                ``temp.x--; ``// Move back to current cell`
`            ``}`
`            ``// Check left adjacent cell that if it can be rotten`
`            ``if` `(isvalid(temp.x-1, temp.y) && arr[temp.x-1][temp.y] == 1) {`
`                ``if` `(!flag) ans++, flag = ``true``;`
`                ``arr[temp.x-1][temp.y] = 2;`
`                ``temp.x--;`
`                ``Q.push(temp); ``// push this cell to Queue`
`                ``temp.x++;`
`            ``}`
`            ``// Check top adjacent cell that if it can be rotten`
`            ``if` `(isvalid(temp.x, temp.y+1) && arr[temp.x][temp.y+1] == 1) {`
`                ``if` `(!flag) ans++, flag = ``true``;`
`                ``arr[temp.x][temp.y+1] = 2;`
`                ``temp.y++;`
`                ``Q.push(temp); ``// Push this cell to Queue`
`                ``temp.y--;`
`            ``}`
`            ``// Check bottom adjacent cell if it can be rotten`
`            ``if` `(isvalid(temp.x, temp.y-1) && arr[temp.x][temp.y-1] == 1) {`
`                ``if` `(!flag) ans++, flag = ``true``;`
`                ``arr[temp.x][temp.y-1] = 2;`
`                ``temp.y--;`
`                ``Q.push(temp); ``// push this cell to Queue`
`            ``}`
`            ``Q.pop();`
`        ``}`
`        ``// Pop the delimiter`
`        ``Q.pop();`
`        ``// If oranges were rotten in current frame than separate the`
`        ``// rotten oranges using delimiter for the next frame for processing.`
`        ``if` `(!Q.empty()) {`
`            ``temp.x = -1;`
`            ``temp.y = -1;`
`            ``Q.push(temp);`
`        ``}`
`        ``// If Queue was empty than no rotten oranges left to process so exit`
`    ``}`
`    ``// Return -1 if all arranges could not rot, otherwise -1.`
`    ``return` `(checkall(arr))? -1: ans;`
`}`
`// Drive program`
`int` `main()`
`{`
`    ``int` `arr[][C] = { {2, 1, 0, 2, 1},`
`                     ``{1, 0, 1, 2, 1},`
`                     ``{1, 0, 0, 2, 1}};`
`    ``int` `ans = rotOranges(arr);`
`    ``if` `(ans == -1)`
`        ``cout << ``"All oranges cannot rotn"``;`
`    ``else`
`         ``cout << ``"Time required for all oranges to rot => "` `<< ans << endl;`
`    ``return` `0;`
`}`

Output:

```Time required for all oranges to rot => 2

```

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
We have built the accelerating growth-oriented website for budding engineers and aspiring job holders of technology companies such as Google, Facebook, and Amazon
If you would like to study our free courses you can join us at