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# Minimum time required to rot all oranges

• Difficulty Level : Hard
• Last Updated : 01 Apr, 2021

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:

## C++14

 `// C++ program to rot all oranges when u can move``// in all the four direction from a rotten orange``#include ``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;``}`

## Java

 `// Java program to rot all oranges when u can move``// in all the four direction from a rotten orange``class` `GFG{``    ` `static` `int` `R = ``3``;``static` `int` `C = ``5``;`` ` `// Check if i, j is under the array``// limits of row and column``static` `boolean` `issafe(``int` `i, ``int` `j)``{``    ``if` `(i >= ``0` `&& i < R &&``        ``j >= ``0` `&& j < C)``        ``return` `true``;``        ` `    ``return` `false``;``}`` ` `static` `int` `rotOranges(``int` `v[][])``{``    ``boolean` `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 Code``public` `static` `void` `main(String[] args)``{``    ``int` `v[][] = { { ``2``, ``1``, ``0``, ``2``, ``1` `},``                  ``{ ``1``, ``0``, ``1``, ``2``, ``1` `},``                  ``{ ``1``, ``0``, ``0``, ``2``, ``1` `} };`` ` `    ``System.out.println(``"Max time incurred: "` `+``                       ``rotOranges(v));``}``}` `// This code is contributed by divyesh072019`

## Python3

 `# 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`

## C#

 `// C# program to rot all oranges when u can move``// in all the four direction from a rotten orange``using` `System;``class` `GFG {``    ` `    ``static` `int` `R = 3;``    ``static` `int` `C = 5;``     ` `    ``// Check if i, j is under the array``   ``// limits of row and column``    ``static` `bool` `issafe(``int` `i, ``int` `j)``    ``{``        ``if` `(i >= 0 && i < R && j >= 0 && j < C)``            ``return` `true``;``        ``return` `false``;``    ``}``     ` `    ``static` `int` `rotOranges(``int``[,] v)``    ``{``        ``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;``    ``}` `  ``static` `void` `Main() {``      ` `    ``int``[ , ] v = { { 2, 1, 0, 2, 1 },``                ``{ 1, 0, 1, 2, 1 },``                ``{ 1, 0, 0, 2, 1 } };`` ` `    ``Console.Write(``"Max time incurred: "` `+ rotOranges(v));``  ``}``}` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``
Output:

`Max time incurred: 2`

• Complexity Analysis:
• Time Complexity : O((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)/2 times. So time complexity is O((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++

 `// C++ program to find minimum time required to make all``// oranges rotten``#include``#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 Q;``    ``ele temp;``    ``int` `ans = 0;` `    ``// Store all the cells having rotten orange in first time frame``    ``for` `(``int` `i=0; i "` `<< 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 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 ans``        ``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`

## Python3

 `# Python3 program to find minimum time required to make all``# oranges rotten``from` `collections ``import` `deque` `# function to check whether a cell is valid / invalid``def` `isvalid(i, j):``    ``return` `(i >``=` `0` `and` `j >``=` `0` `and` `i < ``3` `and` `j < ``5``)` `# Function to check whether the cell is delimiter``# which is (-1, -1)``def` `isdelim(temp):``    ``return` `(temp[``0``] ``=``=` `-``1` `and` `temp[``1``] ``=``=` `-``1``)` `# Function to check whether there is still a fresh``# orange remaining``def` `checkall(arr):``    ``for` `i ``in` `range``(``3``):``       ``for` `j ``in` `range``(``5``):``          ``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``def` `rotOranges(arr):``  ` `    ``# Create a queue of cells``    ``Q ``=` `deque()``    ``temp ``=` `[``0``, ``0``]``    ``ans ``=` `1` `    ``# Store all the cells having``    ``# rotten orange in first time frame``    ``for` `i ``in` `range``(``3``):``       ``for` `j ``in` `range``(``5``):``            ``if` `(arr[i][j] ``=``=` `2``):``                ``temp[``0``]``=` `i``                ``temp[``1``] ``=` `j``                ``Q.append([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)``    ``temp[``0``] ``=` `-``1``    ``temp[``1``] ``=` `-``1``    ``Q.append([``-``1``, ``-``1``])``    ``# print(Q)` `    ``# Process the grid while there are``    ``# rotten oranges in the Queue``    ``while` `False``:``      ` `        ``# 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.``        ``flag ``=` `False``        ``print``(``len``(Q))` `        ``# Process all the rotten``        ``# oranges in current time frame.``        ``while` `not` `isdelim(Q[``0``]):``            ``temp ``=` `Q[``0``]``            ``print``(``len``(Q))` `            ``# Check right adjacent cell that if it can be rotten``            ``if` `(isvalid(temp[``0``] ``+` `1``, temp[``1``]) ``and` `arr[temp[``0``] ``+` `1``][temp[``1``]] ``=``=` `1``):``                ` `                ``# if this is the first orange to get rotten, increase``                ``# count and set the flag.``                ``if` `(``not` `flag):``                    ``ans, flag ``=``ans ``+` `1``, ``True` `                ``# Make the orange rotten``                ``arr[temp[``0``] ``+` `1``][temp[``1``]] ``=` `2` `                ``# append the adjacent orange to Queue``                ``temp[``0``] ``+``=` `1``                ``Q.append(temp)` `                ``temp[``0``] ``-``=` `1` `# Move back to current cell` `            ``# Check left adjacent cell that if it can be rotten``            ``if` `(isvalid(temp[``0``] ``-` `1``, temp[``1``]) ``and` `arr[temp[``0``] ``-` `1``][temp[``1``]] ``=``=` `1``):``                ``if` `(``not` `flag):``                    ``ans, flag ``=``ans ``+` `1``, ``True``                ``arr[temp[``0``] ``-` `1``][temp[``1``]] ``=` `2``                ``temp[``0``] ``-``=` `1``                ``Q.append(temp) ``# append this cell to Queue``                ``temp[``0``] ``+``=` `1` `            ``# Check top adjacent cell that if it can be rotten``            ``if` `(isvalid(temp[``0``], temp[``1``] ``+` `1``) ``and` `arr[temp[``0``]][temp[``1``] ``+` `1``] ``=``=` `1``):``                ``if` `(``not` `flag):``                    ``ans, flag ``=` `ans ``+` `1``, ``True``                ``arr[temp[``0``]][temp[``1``] ``+` `1``] ``=` `2``                ``temp[``1``] ``+``=` `1``                ``Q.append(temp) ``# Push this cell to Queue``                ``temp[``1``] ``-``=` `1` `            ``# Check bottom adjacent cell if it can be rotten``            ``if` `(isvalid(temp[``0``], temp[``1``] ``-` `1``) ``and` `arr[temp[``0``]][temp[``1``] ``-` `1``] ``=``=` `1``):``                ``if` `(``not` `flag):``                    ``ans, flag ``=` `ans ``+` `1``, ``True``                ``arr[temp[``0``]][temp[``1``] ``-` `1``] ``=` `2``                ``temp[``1``] ``-``=` `1``                ``Q.append(temp) ``# append this cell to Queue``            ``Q.popleft()` `        ``# Pop the delimiter``        ``Q.popleft()` `        ``# If oranges were rotten in``        ``# current frame than separate the``        ``# rotten oranges using delimiter``        ``# for the next frame for processing.``        ``if` `(``len``(Q) ``=``=` `0``):``            ``temp[``0``] ``=` `-``1``            ``temp[``1``] ``=` `-``1``            ``Q.append(temp)` `        ``# If Queue was empty than no rotten oranges left to process so exit` `    ``# Return -1 if all arranges could not rot, otherwise return ans.``    ``return` `ans ``+` `1` `if``(checkall(arr)) ``else` `-``1` `# Driver program``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[[``2``, ``1``, ``0``, ``2``, ``1``],``         ``[``1``, ``0``, ``1``, ``2``, ``1``],``         ``[``1``, ``0``, ``0``, ``2``, ``1``]]``    ``ans ``=` `rotOranges(arr)``    ``if` `(ans ``=``=` `-``1``):``        ``print``(``"All oranges cannot rotn"``)``    ``else``:``        ``print``(``"Time required for all oranges to rot => "` `, ans)` `        ``# This code is contributed by mohit kumar 29`

## 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 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.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 ans``        ``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 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.