994. Rotting Oranges

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. 1 <= grid.length <= 10
2. 1 <= grid[0].length <= 10
3. grid[i][j] is only 0, 1, or 2.

# @lc code=start
using LeetCode, DataStructures

function oranges_rotting(grid::Vector{Vector{Int}})
    nr, nc = length(grid), length(grid[1])
    visited = fill(false, nr, nc)
    q = Queue{Tuple{Int,Int,Int}}()
    left, res = 0, 0
    for r in 1:nr, c in 1:nc
        if grid[r][c] == 2
            enqueue!(q, (r, c, 0))
        elseif grid[r][c] == 1
            left += 1
        end
    end
    dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
    while !isempty(q)
        tp = dequeue!(q)
        for dir in dirs
            nb = (tp[1] + dir[1], tp[2] + dir[2], tp[3] + 1)
            if 0 < nb[1] ≤ nr &&
               0 < nb[2] ≤ nc &&
               !visited[nb[1], nb[2]] &&
               grid[nb[1]][nb[2]] == 1
                left -= 1
                visited[nb[1], nb[2]] = true
                enqueue!(q, nb)
            end
        end
        res = tp[3]
    end
    left == 0 ? res : -1
end
# @lc code=end

oranges_rotting (generic function with 1 method)

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