```
//my solution
class Solution {
public int orangesRotting(int[][] grid) {
//n2 to find the first rotten tomato
Queue<Indices> q = new LinkedList<>();
for(int i =0;i<grid.length;i++){
for(int j =0;j<grid[0].length;j++){
if(grid[i][j]!=2) continue;
q.add(new Indices(i,j));
}
}
//do breadth first search to set all the tomatos rotten
int timer = 0;
while(!q.isEmpty()){
int size = q.size();
boolean newRotting = false;
while(size-->0){
Indices in = q.remove();
//left
if(in.j-1>=0 && grid[in.i][in.j-1]==1){
grid[in.i][in.j-1] = 2;
newRotting = true;
q.add(new Indices(in.i,in.j-1));
}
//right
if(in.j+1<grid[0].length && grid[in.i][in.j+1] ==1){
grid[in.i][in.j+1] =2;
newRotting = true;
q.add(new Indices(in.i,in.j+1));
}
//up
if(in.i-1>=0 && grid[in.i-1][in.j] ==1){
grid[in.i-1][in.j] = 2;
newRotting = true;
q.add(new Indices(in.i-1,in.j));
}
//down
if(in.i +1 < grid.length && grid[in.i+1][in.j]==1){
grid[in.i+1][in.j] = 2;
newRotting =true;
q.add(new Indices(in.i+1,in.j));
}
}
if(newRotting) timer++;
}
//n^2 to check if there are any tomatos left freash if yes return -1 else return minutes elapsed.
for(int i =0;i<grid.length;i++){
for(int j =0;j<grid[0].length;j++){
if(grid[i][j]==1) return -1;
}
}
for(int i =0;i<grid.length;i++){
for(int j =0;j<grid[0].length;j++){
System.out.print(grid[i][j]);
}
System.out.println();
}
return timer;
}
}
class Indices{
int i;
int j;
public Indices(int i, int j) {
this.i =i;
this.j = j;
}
}
```

```
// second solution from one of the submission that I saw
//great solution
class Solution {
public int orangesRotting(int[][] grid) {
int m=grid.length,n=grid[0].length,i,j,k=0,fresh=0,fr;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
if(grid[i][j]==1) fresh++;
while(fresh>0){
fr=fresh;
for(i=0;i<m;i++){
for(j=0;j<n;j++){
if(grid[i][j]==2){
if(i+1<m && grid[i+1][j]==1) {grid[i+1][j]=3;fresh--;}
if(i-1>=0 && grid[i-1][j]==1) {grid[i-1][j]=3;fresh--;}
if(j-1>=0 && grid[i][j-1]==1) {grid[i][j-1]=3;fresh--;}
if(j+1<n && grid[i][j+1]==1) {grid[i][j+1]=3;fresh--;}
}
}
}
for(i=0;i<m;i++)
for(j=0;j<n;j++)
if(grid[i][j]==3) grid[i][j]=2;
if(fr==fresh) return -1;
k++;
}
return k;
}
}
```

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