Problem Statement
Medium
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, or2 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.lengthn == grid[i].length1 <= m, n <= 10grid[i][j] is 0, 1, or 2.
class Solution {
private static final int[][] dirs = {
{ 0, -1 }, { 0, 1 }, { 1, 0 }, { -1, 0 }
};
public int orangesRotting(int[][] grid) {
Queue<int[]> pq = new LinkedList<>();
int m = grid.length, n = grid[0].length;
int depth = 0, totalCt = 0, rottenCt = 0, nodesAtLevel;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++) {
if (grid[i][j] == 2) {
rottenCt++;
pq.offer(new int[] { i, j });
}
if (grid[i][j] != 0)
totalCt++;
}
nodesAtLevel = pq.size();
if(totalCt==0)
return 0;
while (!pq.isEmpty()) {
int[] top = pq.poll();
grid[top[0]][top[1]] = 0;
// System.out.println(Arrays.toString(top) + "\t" + nodesAtLevel + "\t" + depth);
for (int[] d : dirs) {
int i = top[0] + d[0], j = top[1] + d[1];
if (i < 0 || j < 0 || i >= m || j >= n)
continue;
if (grid[i][j] == 1) {
rottenCt++;
grid[i][j] = 2;
pq.offer(new int[] { i, j });
}
}
if (--nodesAtLevel == 0) {
depth++;
nodesAtLevel = pq.size();
}
}
if (rottenCt < totalCt)
return -1;
return depth - 1;
}
}