1. Description
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, or
- 2 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.
2. Example
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.
3. Constraints
- m == grid.length
- n == grid[i].length
- 1 <= m, n <= 10
- grid[i][j] is 0, 1, or 2.
4. Solutions
Breadth-First Search
m = grid.size(), n = grid.front().size()
Time complexity: O(mn)
Space complexity: O(mn)
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| class Solution {
public:
int orangesRotting(vector<vector<int>> &grid) {
const int m = grid.size(), n = grid.front().size();
int fresh = 0;
queue<pair<int, int>> rotten;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
++fresh;
} else if (grid[i][j] == 2) {
rotten.push({i, j});
}
}
}
array<pair<int, int>, 4> directions{{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}};
int minutes = 0;
while (!rotten.empty() && fresh > 0) {
++minutes;
const int count = rotten.size();
for (int i = 0; i < count; ++i) {
const auto [r, c] = rotten.front();
rotten.pop();
for (const auto &[rd, cd] : directions) {
int next_row = r + rd;
int next_column = c + cd;
if (0 <= next_row && next_row < m && 0 <= next_column && next_column < n &&
grid[next_row][next_column] == 1) {
grid[next_row][next_column] = 2;
--fresh;
if (fresh == 0) {
goto final;
}
rotten.push({next_row, next_column});
}
}
}
}
final:
return fresh == 0 ? minutes : -1;
}
};
|