Formatted question description: https://leetcode.ca/all/994.html

994. Rotting Oranges (Medium)

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.

Related Topics: Breadth-first Search

Similar Questions:

Walls and Gates (Medium)

Solution 1. BFS

// OJ: https://leetcode.com/problems/rotting-oranges/
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int orangesRotting(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
        queue<pair<int, int>> q;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 2) q.emplace(i, j);
            }
        }
        if (q.empty()) ans = 1;
        while (q.size()) {
            int cnt = q.size();
            ++ans;
            while (cnt--) {
                auto [x, y] = q.front();
                q.pop();
                for (auto &dir : dirs) {
                    int a = x + dir[0], b = y + dir[1];
                    if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] != 1) continue;
                    A[a][b] = 2;
                    q.emplace(a, b);
                }
            }
        }
        for (auto &row : A) {
            for (int x : row) {
                if (x == 1) return -1;
            }
        }
        return ans - 1;
    }
};

Java

class Solution {
    public int orangesRotting(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0)
            return 0;
        int rows = grid.length, columns = grid[0].length;
        final int BLOCK = -1;
        final int WHITE = 0;
        final int GRAY = 1;
        final int BLACK = 2;
        int[][] colors = new int[rows][columns];
        int[][] distances = new int[rows][columns];
        Queue<int[]> queue = new LinkedList<int[]>();
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                int value = grid[i][j];
                if (value == 2) {
                    distances[i][j] = 0;
                    colors[i][j] = GRAY;
                    queue.offer(new int[]{i, j});
                } else if (value == 1) {
                    distances[i][j] = Integer.MAX_VALUE;
                    colors[i][j] = WHITE;
                } else {
                    distances[i][j] = -1;
                    colors[i][j] = BLOCK;
                }
            }
        }
        int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int row = cell[0], column = cell[1];
            int distance = distances[row][column];
            for (int[] direction : directions) {
                int newRow = row + direction[0], newColumn = column + direction[1];
                if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns) {
                    if (colors[newRow][newColumn] == WHITE) {
                        distances[newRow][newColumn] = distance + 1;
                        colors[newRow][newColumn] = GRAY;
                        queue.offer(new int[]{newRow, newColumn});
                    }
                }
            }
            colors[row][column] = BLACK;
        }
        int time = 0;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                time = Math.max(time, distances[i][j]);
                if (time == Integer.MAX_VALUE)
                    return -1;
            }
        }
        return time == Integer.MAX_VALUE ? -1 : time;
    }
}

// OJ: https://leetcode.com/problems/rotting-oranges/
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int orangesRotting(vector<vector<int>>& A) {
        queue<pair<int, int>> q;
        int M = A.size(), N = A[0].size(), step = 0, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 2) q.emplace(i, j);
            }
        }
        while (q.size()) {
            for (int cnt = q.size(); cnt--;) {
                auto [x, y] = q.front();
                q.pop();
                for (auto &[dx, dy] : dirs) {
                    int a = x + dx, b = y + dy;
                    if (a < 0 || a >= M || b < 0 || b >= N || A[a][b] != 1) continue;
                    A[a][b] = 2;
                    q.emplace(a, b);
                }
            }
            step++;
        }
        for (auto &row : A) {
            for (int x : row) {
                if (x == 1) return -1;
            }
        }
        return max(0, step - 1);
    }
};

class Solution(object):
    def orangesRotting(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        M, N = len(grid), len(grid[0])
        fresh = 0
        q = collections.deque()
        for i in range(M):
            for j in range(N):
                if grid[i][j] == 1:
                    fresh += 1
                elif grid[i][j] == 2:
                    q.append((i, j))
        if fresh == 0:
            return 0
        dirs = [(0, 1), (0, -1), (-1, 0), (1, 0)]
        step = 0
        while q:
            size = len(q)
            for i in range(size):
                x, y = q.popleft()
                for d in dirs:
                    nx, ny = x + d[0], y + d[1]
                    if nx < 0 or nx >= M or ny < 0 or ny >= N or grid[nx][ny] != 1:
                        continue
                    grid[nx][ny] = 2
                    q.append((nx, ny))
                    fresh -= 1
            step += 1
        if fresh != 0:
            return -1
        return step - 1

994 - Rotting Oranges

994. Rotting Oranges (Medium)

Solution 1. BFS

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