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: 
     

CPP14

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// 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;
}

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Python3

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# Python3 program to rot all
# oranges when u can move
# in all the four direction
# from a rotten orang
R = 3
C = 5
 
# Check if i, j is under the
# array limits of row and
# column
def issafe(i, j):
 
    if (i >= 0 and i < R and
        j >= 0 and j < C):
        return True
    return False
 
def rotOranges(v):
 
    changed = False
    no = 2
    while (True):
        for i in range(R):
            for j in range(C):
 
                # 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) and
                        v[i + 1][j] == 1):
                        v[i + 1][j] = v[i][j] + 1
                        changed = True
 
                    if (issafe(i, j + 1) and
                        v[i][j + 1] == 1):
                        v[i][j + 1] = v[i][j] + 1
                        changed = True
 
                    if (issafe(i - 1, j) and
                        v[i - 1][j] == 1):
                        v[i - 1][j] = v[i][j] + 1
                        changed = True
 
                    if (issafe(i, j - 1) and
                        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 (not changed):
           break
        changed = False
        no += 1
 
    for i in range(R):
        for j in range(C):
 
            # 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
if __name__ == "__main__":
 
    v = [[2, 1, 0, 2, 1],
         [1, 0, 1, 2, 1],
         [1, 0, 0, 2, 1]]
 
    print("Max time incurred: ",
           rotOranges(v))
 
# This code is contributed by Chitranayal

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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 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: 
     

C++

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// 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;
}

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//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

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C#

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// 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

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  • 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 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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