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 only0
,1
, or2
.
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
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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; } }
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// 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); } };
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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