994. Rotting Oranges

2023, Aug 27 2 mins read

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

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)

class Solution {
public:
    int orangesRotting(vector<vector<int>> grid) {
        int fresh_count = 0;
        const int m = grid.size(), n = grid.front().size();
        queue<pair<int, int>> rotten_oranges;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    ++fresh_count;
                } else if (grid[i][j] == 2) {
                    rotten_oranges.push(make_pair(i, j));
                }
            }
        }

        if (fresh_count == 0) {
            return 0;
        }

        int step = 0;
        const array<pair<int, int>, 4> dirs{{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}};
        while (!rotten_oranges.empty()) {
            ++step;
            for (int i = 0, count = rotten_oranges.size(); i < count; ++i) {
                auto rotten = rotten_oranges.front();
                rotten_oranges.pop();

                for (auto dir : dirs) {
                    if (rotten.first + dir.first >= 0 && rotten.first + dir.first < m &&
                        rotten.second + dir.second >= 0 && rotten.second + dir.second < n &&
                        grid[rotten.first + dir.first][rotten.second + dir.second] == 1) {
                        grid[rotten.first + dir.first][rotten.second + dir.second] = 2;
                        rotten_oranges.push(
                            make_pair(rotten.first + dir.first, rotten.second + dir.second));

                        --fresh_count;
                        if (fresh_count == 0) {
                            return step;
                        }
                    }
                }
            }
        }

        return -1;
    }
};