# Solution For Rotting Oranges

The Rotting Oranges problem on LeetCode asks you to solve a problem where you are given a 2D grid representing the initial status of a bunch of oranges, with values of 0 representing an empty cell, 1 representing a fresh orange, and 2 representing a rotten orange. Each minute, any fresh orange that is adjacent to a rotten orange becomes rotten too. Determine the minimum number of minutes that must elapse until no cell has a fresh orange. If this is not possible, return -1 instead.

To solve this problem, we can use BFS (breadth-first search) to simulate the process of the oranges getting rotten. First, we can count the number of fresh oranges and the positions of all the rotten oranges. Then, we can start BFS from the positions of all the rotten oranges, and for each level of the BFS, all the fresh oranges that are adjacent to any of the rotten oranges will become rotten. We can keep track of the number of fresh oranges that are turned into rotten oranges in each level of the BFS, and add this number to the total number of minutes it takes to fully rot all the fresh oranges.

If all the fresh oranges have been rotten at the end, return the total number of minutes. Otherwise, return -1.

Here is the detailed solution in Python:

“`
from collections import deque

class Solution:
def orangesRotting(self, grid: List[List[int]]) -> int:
# find all the positions of the rotten oranges
rotten = [] for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 2:
rotten.append((i, j))

``````    # BFS to simulate the process of the oranges getting rotten
fresh_count = sum(1 for row in grid for cell in row if cell == 1)
minutes = 0
while rotten:
new_rotten = []
for i, j in rotten:
for x, y in [(i+1,j),(i-1,j),(i,j+1),(i,j-1)]:
if 0 <= x < len(grid) and 0 <= y < len(grid[0]) and grid[x][y] == 1:
grid[x][y] = 2
fresh_count -= 1
new_rotten.append((x, y))
rotten = new_rotten
if rotten:
minutes += 1

return minutes if fresh_count == 0 else -1
``````

“`

Time Complexity: O(N), where N is the total number of cells in the grid, since we visit each cell at most once in the BFS.

Space Complexity: O(N), for the queue used in the BFS and the list used to store the positions of the rotten oranges.

## Step by Step Implementation For Rotting Oranges

```There are a number of ways to solve this problem. One approach would be to use a breadth-first search algorithm to find all of the oranges that are currently rotten and then to keep track of how long it takes for each orange to rot. Once all of the oranges have been rotted, the maximum time it took for any orange to rot can be returned as the answer.

Another approach would be to use a depth-first search algorithm to find all of the oranges that are currently rotten and then to keep track of how long it takes for each orange to rot. Once all of the oranges have been rotted, the maximum time it took for any orange to rot can be returned as the answer.

A third approach would be to use a union-find algorithm to keep track of which oranges are connected to which other oranges. Once all of the oranges have been rotted, the maximum time it took for any orange to rot can be returned as the answer.```
`There is no one-size-fits-all answer to this question, as the solution to the leetcode problem rotting-oranges will vary depending on the specific problem. However, a potential Python solution to this problem could involve using a queue to keep track of the oranges that need to be rotten, and then using a breadth-first search to determine the minimum number of minutes needed to rot all the oranges.`
```There are a number of ways to solve this problem. One approach would be to use a breadth-first search algorithm to find all of the oranges that are rotten and then to mark them as rotten. This approach would be similar to the one used in the leetcode problem Asynchronous JavaScript.

Another approach would be to use a depth-first search algorithm to find all of the oranges that are rotten and then to mark them as rotten. This approach would be similar to the one used in the leetcode problem Rotting Oranges II.

Both of these approaches would require the use of a queue or stack in order to keep track of the oranges that need to be processed.```
```There are a total of n oranges in the grid, n rows and n columns.

Initially, some of the oranges are rotten and some are fresh.

Every orange that is adjacent (4-directionally) to a rotten orange becomes rotten.

Return the minimum number of oranges that must be rotten so that every orange in the grid is rotten.

If it is impossible to rot every orange, return -1 instead.

class Solution {
public:
int orangesRotting(vector>& grid) {
int n = grid.size();
int m = grid[0].size();
queue> q;
int fresh = 0;
for(int i=0;i= n || ny < 0 || ny >= m || grid[nx][ny] == 0 || grid[nx][ny] == 2) continue;
grid[nx][ny] = 2;
fresh--;
q.push({nx, ny});
flag = true;
}
}
if(flag) ans++;
}
return fresh == 0 ? ans : -1;
}
};```
```There are a few things we need to do in order to solve this problem. First, we need to create a class that represents an Orange. This class will have a few properties that will help us keep track of the state of each Orange. Next, we need to create a method that will take in an array of Oranges and return the number of minutes it will take for all of the Oranges to rot.

class Orange { public int Row { get; set; } public int Column { get; set; } public int MinutesUntilRot { get; set; } public Orange(int row, int column, int minutesUntilRot) { Row = row; Column = column; MinutesUntilRot = minutesUntilRot; } } int RottingOranges(Orange[,] oranges) { // TODO: Implement solution here }```

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