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1036. Rotting Oranges

Array Breadth-First Search Matrix

Problem - Rotting Oranges

Medium

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.

Solutions

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class Solution:
    def orangesRotting(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        count = 0
        queue = deque()

        for i, row in enumerate(grid):
            for j, val in enumerate(row):
                if val == 2:
                    queue.append((i, j))
                elif val == 1:
                    count += 1

        result = 0
        directions = (-1, 0, 1, 0, -1)
        while queue and count:
            result += 1
            for _ in range(len(queue)):
                i, j = queue.popleft()
                for a, b in pairwise(directions):
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n and grid[x][y] == 1:
                        grid[x][y] = 2
                        queue.append((x, y))
                        count -= 1
                        if count == 0:
                            return result

        return -1 if count else 0

Submission Stats:

  • Runtime: 3 ms (76.53%)
  • Memory: 19.3 MB (41.20%)