# 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.

```

Source: Microsoft Interview Question

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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++

```// 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;
}
```

## Java

```//Java program to find minimum time required to make all
//oranges rotten

import java.util.Queue;

public class RotOrange
{
public final static int R = 3;
public final static int C = 5;

// structure for storing coordinates of the cell
static class Ele
{
int x = 0;
int y = 0;
Ele(int x,int y)
{
this.x = x;
this.y = y;
}
}

// function to check whether a cell is valid / invalid
static boolean isValid(int i, int j)
{
return (i >= 0 && j >= 0 && i < R && j < C);
}

// Function to check whether the cell is delimiter
// which is (-1, -1)
static boolean isDelim(Ele temp)
{
return (temp.x == -1 && temp.y == -1);
}

// Function to check whether there is still a fresh
// orange remaining
static boolean checkAll(int arr[][])
{
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
static int rotOranges(int arr[][])
{
// Create a queue of cells
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)

// Separate these rotten oranges from the oranges which will rotten
// due the oranges in first time frame using delimiter which is (-1, -1)

// Process the grid while there are rotten oranges in the Queue
while(!Q.isEmpty())
{
// 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.
boolean flag = false;

// Process all the rotten oranges in current time frame.
while(!isDelim(Q.peek()))
{
temp = Q.peek();

// Check right adjacent cell that if it can be rotten
if(isValid(temp.x+1, temp.y+1) && arr[temp.x+1][temp.y] == 1)
{
if(!flag)
{
// if this is the first orange to get rotten, increase
// count and set the flag.
ans++;
flag = true;
}
// Make the orange rotten
arr[temp.x+1][temp.y] = 2;

// push the adjacent orange to Queue
temp.x++;

// Move back to current cell
temp.x--;
}

// 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.add(new Ele(temp.x,temp.y)); // 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.add(new Ele(temp.x,temp.y)); // 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.add(new Ele(temp.x,temp.y)); // push this cell to Queue
}
Q.remove();

}
// Pop the delimiter
Q.remove();

// If oranges were rotten in current frame than separate the
// rotten oranges using delimiter for the next frame for processing.
if (!Q.isEmpty())
{
}

// If Queue was empty than no rotten oranges left to process so exit
}

// Return -1 if all arranges could not rot, otherwise -1.s
return (checkAll(arr))? -1: ans;

}

// Drive program
public static void main(String[] args)
{
int arr[][] = { {2, 1, 0, 2, 1},
{1, 0, 1, 2, 1},
{1, 0, 0, 2, 1}};
int ans = rotOranges(arr);
if(ans == -1)
System.out.println("All oranges cannot rot");
else
System.out.println("Time required for all oranges to rot = " + ans);
}

}
//This code is contributed by Sumit Ghosh
```

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

Thanks to Gaurav Ahirwar for suggesting above solution here

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.7 Average Difficulty : 3.7/5.0
Based on 146 vote(s)