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

Related Topics: Breadth-first Search

Similar Questions:

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
    

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