994. Rotting Oranges

#### QUESTION:

In a given grid, each cell can have one of three values:

the value 0 representing an empty cell; the value 1 representing a fresh orange; the value 2 representing a rotten orange. Every minute, any fresh orange that is adjacent (4-directionally) 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 instead.

Example 1: Input: [[2,1,1],[1,1,0],[0,1,1]] Output: 4

Example 2:

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

Note:

1 <= grid.length <= 10 1 <= grid[0].length <= 10 grid[i][j] is only 0, 1, or 2.

#### SOLUTION:

``````class Solution {
public int orangesRotting(int[][] grid) {
Queue<int[]> queue = new ArrayDeque<>();
int count = 0;
while (true){
for(int i = 0;i<grid.length;i++){
for(int j = 0;j<grid[0].length;j++){
if(grid[i][j]==1){
}
}
}
if(queue.isEmpty()) break;
while (!queue.isEmpty()){
int[] poll = queue.poll();
grid[poll[0]][poll[1]] = 2;
}
count++;
}
if(orangesRottingHelper(grid)) return count;
else return -1;
}

public static boolean orangesRottingHelper(int[][] grid){
for(int i = 0;i<grid.length;i++){
for(int j = 0;j<grid[i].length;j++){
if(grid[i][j]==1) return false;
}
}
return true;
}
}
``````
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