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.

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

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 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 a 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 elements value is equal to 1, i.e. not rotten and update changed to true.
4. Traverse the matrix and check if there are any cell which is 1. If 1 is present return -1
5. Else return no – 2

Implementation:

// C++ program to rot all oranges when u can move 
// in all the four direction from a rotten orange 
#include <bits/stdc++.h> 
using namespace std; 
  
const int R = 3; 
const int C = 5; 
  
// Check if i, j is under the array limits of row and column 
bool issafe(int i, int j) 
{ 
    if (i >= 0 && i < R && j >= 0 && j < C) 
        return true; 
    return false; 
} 
  
int rotOranges(int v[R][C]) 
{ 
    bool changed = false; 
    int no = 2; 
    while (true) { 
        for (int i = 0; i < R; i++) { 
            for (int j = 0; j < C; j++) { 
  
                // Rot all other oranges present at 
                // (i+1, j), (i, j-1), (i, j+1), (i-1, j) 
                if (v[i][j] == no) { 
                    if (issafe(i + 1, j) && v[i + 1][j] == 1) { 
                        v[i + 1][j] = v[i][j] + 1; 
                        changed = true; 
                    } 
                    if (issafe(i, j + 1) && v[i][j + 1] == 1) { 
                        v[i][j + 1] = v[i][j] + 1; 
                        changed = true; 
                    } 
                    if (issafe(i - 1, j) && v[i - 1][j] == 1) { 
                        v[i - 1][j] = v[i][j] + 1; 
                        changed = true; 
                    } 
                    if (issafe(i, j - 1) && v[i][j - 1] == 1) { 
                        v[i][j - 1] = v[i][j] + 1; 
                        changed = true; 
                    } 
                } 
            } 
        } 
  
        // if no rotten orange found it means all 
        // oranges rottened now 
        if (!changed) 
            break; 
        changed = false; 
        no++; 
    } 
  
    for (int i = 0; i < R; i++) { 
        for (int j = 0; j < C; j++) { 
  
            // if any orange is found to be 
            // not rotten then ans is not possible 
            if (v[i][j] == 1) 
                return -1; 
        } 
    } 
  
    // Because initial value for a rotten 
    // orange was 2 
    return no - 2; 
} 
  
// Driver function 
int main() 
{ 
  
    int v[R][C] = { { 2, 1, 0, 2, 1 }, 
                    { 1, 0, 1, 2, 1 }, 
                    { 1, 0, 0, 2, 1 } }; 
  
    cout << "Max time incurred: " << rotOranges(v); 
  
    return 0; 
} 

Output:

Max time incurred: 2

Complexity Analysis:
- Time Complexity : O(max(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) times. So time complexity is O(max(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 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:

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; 
} 
  
// Driver 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.LinkedList; 
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 
        Queue<Ele> Q=new LinkedList<>(); 
        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) 
                   Q.add(new Ele(i,j)); 
                  
        // Separate these rotten oranges from the oranges which will rotten 
        // due the oranges in first time frame using delimiter which is (-1, -1) 
        Q.add(new Ele(-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 the 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) && 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++; 
                    Q.add(new Ele(temp.x,temp.y)); 
                      
                    // 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())  
            { 
                Q.add(new Ele(-1,-1)); 
            } 
              
            // 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; 
          
    } 
      
    // Driver 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 

C#

// C# program to find minimum time  
// required to make all oranges rotten  
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
    public const int R = 3; 
    public const int C = 5; 
  
    // structure for storing  
    // coordinates of the cell  
    public class Ele 
    { 
        public int x = 0; 
        public int y = 0; 
        public Ele(int x, int y) 
        { 
            this.x = x; 
            this.y = y; 
        } 
    } 
  
    // function to check whether a cell 
    // is valid / invalid  
    public static bool 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)  
    public static bool isDelim(Ele temp) 
    { 
        return (temp.x == -1 && temp.y == -1); 
    } 
  
    // Function to check whether there  
    // is still a fresh orange remaining  
    public static bool 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  
    public static int rotOranges(int[][] arr) 
    { 
        // Create a queue of cells  
        LinkedList<Ele> Q = new LinkedList<Ele>(); 
        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) 
            { 
                Q.AddLast(new Ele(i, j)); 
            } 
        } 
        } 
  
        // Separate these rotten oranges from  
        // the oranges which will rotten  
        // due the oranges in first time frame 
        // using delimiter which is (-1, -1)  
        Q.AddLast(new Ele(-1,-1)); 
  
        // Process the grid while there are  
        // rotten oranges in the Queue  
        while (Q.Count > 0) 
        { 
            // 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.First.Value)) 
            { 
                temp = Q.First.Value; 
  
                // 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 (!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++; 
                    Q.AddLast(new Ele(temp.x,temp.y)); 
  
                    // 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--; 
                          
                        // push this cell to Queue 
                        Q.AddLast(new Ele(temp.x,temp.y));  
                        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++; 
                          
                        // Push this cell to Queue 
                        Q.AddLast(new Ele(temp.x,temp.y));  
                        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--; 
                          
                        // push this cell to Queue 
                        Q.AddLast(new Ele(temp.x,temp.y));  
                } 
                Q.RemoveFirst(); 
  
            } 
              
            // Pop the delimiter  
            Q.RemoveFirst(); 
  
            // If oranges were rotten in current  
            // frame than separate the rotten  
            // oranges using delimiter for the  
            // next frame for processing.  
            if (Q.Count > 0) 
            { 
                Q.AddLast(new Ele(-1,-1)); 
            } 
  
            // 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; 
  
    } 
  
    // Driver Code  
    public static void Main(string[] args) 
    { 
        int[][] arr = new int[][] 
        { 
            new int[] {2, 1, 0, 2, 1}, 
            new int[] {1, 0, 1, 2, 1}, 
            new int[] {1, 0, 0, 2, 1} 
        }; 
          
        int ans = rotOranges(arr); 
        if (ans == -1) 
        { 
            Console.WriteLine("All oranges cannot rot"); 
        } 
        else
        { 
            Console.WriteLine("Time required for all " + 
                              "oranges to rot => " + ans); 
        } 
    } 
} 
  
// This code is contributed by Shrikant13 

Output:

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 upperbound 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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My Personal Notes

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

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