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.
Solution
class Solution {
public:
int orangesRotting(vector<vector<int>> &grid) {
int day = 0, fresh = 0;
queue<array<int, 2>> q;
for (int i = 0; i < grid.size(); ++i)
for (int j = 0; j < grid[0].size(); ++j)
if (grid[i][j] == 1)
++fresh;
else if (grid[i][j] == 2)
q.push({i, j});
vector<array<int, 2>> dirs = {
{-1, 0}, {1, 0}, {0, -1}, {0, 1}
};
while (!q.empty()) {
int n = q.size();
bool rotten = false;
for (int i = 0; i < n; ++i) {
auto x = q.front();
q.pop();
for (auto dir : dirs) {
int i = x[0] + dir[0];
int j = x[1] + dir[1];
if (0 <= i && i < grid.size() && 0 <= j && j < grid[0].size() && grid[i][j] == 1) {
grid[i][j] = 2;
q.push({i, j});
--fresh;
rotten = true;
}
}
}
if (rotten)
++day;
}
return fresh ? -1 : day;
}
};
