leetcode/work/1/Medium/Array/994.rotting-oranges.py

"""
994. Rotting Oranges
Difficulty: Medium
https://leetcode.com/problems/rotting-oranges/

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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.



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.



Constraints:

        • m == grid.length

        • n == grid[i].length

        • 1 <= m, n <= 10

        • grid[i][j] is 0, 1, or 2.
"""

from collections import deque


class Solution:
    def orangesRotting(self, grid: list[list[int]]) -> int:
        q = deque()
        time, fresh = 0, 0

        rows, cols = len(grid), len(grid[0])
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == 1:
                    fresh += 1
                if grid[r][c] == 2:
                    q.append([r, c])

        directions = [[0, 1], [0, -1], [1, 0], [-1, 0]]

        while q and fresh > 0:
            for i in range(len(q)):
                r, c = q.popleft()
                for dr, dc in directions:
                    row, col = dr + r, dc + c
                    # check if in bounds
                    if (
                        row < 0
                        or col < 0
                        or row == len(grid)
                        or col == len(grid[0])
                        or grid[row][col] != 1
                    ):
                        continue
                    grid[row][col] = 2
                    q.append([row, col])
                    fresh -= 1
            time += 1
        return time if fresh == 0 else -1