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)

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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;
    }
};