Rotting oranges

//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; } }