### LeetCode Challenge #994. Rotting Oranges

You are given an `m x n` `grid` where each cell can have one of three values:

• `0` representing an empty cell,
• `1` representing a fresh orange, or
• `2` representing a rotten orange.

Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.

Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return `-1`.

Example 1:

```Input: grid = [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
```

Example 2:

```Input: grid = [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
```

Example 3:

```Input: grid = [[0,2]]
Output: 0
Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
```

Constraints:

• `m == grid.length`
• `n == grid[i].length`
• `1 <= m, n <= 10`
• `grid[i][j]` is `0``1`, or `2`.
##### Video Solution
###### C++ Solution
```				```
class Solution {
public:
int bfs(vector<vector<int>>& grid, int n, int m)
{
queue<pair<int,int>>q;
int count=0;

for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
if(grid[i][j]==2)q.push({i,j});
}
}

while(!q.empty())
{
int size=q.size();
bool isSpoilt=false;

while(size--)
{
int thisRow=q.front().first;
int thisCol=q.front().second;
q.pop();

int x[4]={-1,1,0,0};
int y[4]={0,0,-1,1};

for(int i=0; i<4; i++)
{
int row=thisRow+x[i];
int col=thisCol+y[i];

if(row>=0 and col>=0 and row<=n-1 and col<=m-1 and grid[row][col]==1)
{
grid[row][col]=2;
q.push({row,col});
isSpoilt=true;
}
}
}
if(isSpoilt)count++;
}

return count;

}
int orangesRotting(vector<vector<int>>& grid) {

int n=grid.size();
int m=grid[0].size();

int ans=bfs(grid,n,m);

for(auto i:grid)
{
for(auto j:i)
{
if(j==1)return -1;
}
}

return ans;
}
};
```
```