Rotting Oranges

Clarify the problem:

• The problem requires us to determine the minimum number of minutes required to rot all the fresh oranges in a grid.
• We need to implement a function that takes in the grid as input and returns the minimum number of minutes, or -1 if it is not possible to rot all the oranges.

Analyze the problem:

• Input: 2D grid of integers representing the state of oranges.
• Output: Minimum number of minutes required or -1 if not possible.
• Constraints:
  • The grid can have 0, 1, or 2 representing empty cells, fresh oranges, and rotten oranges respectively.
  • Oranges can rot up, down, left, or right.
  • Rotting happens simultaneously.
  • If all the fresh oranges cannot be rotten, return -1.
  • The grid size is between 1 and 10^3.

Design an algorithm:

• We can solve this problem using a breadth-first search (BFS) approach.
• The idea is to start with the initial state and simulate the rotting process step by step.
• We can use a queue to keep track of the current rotten oranges and their respective minutes.
• We initialize the queue with the coordinates of the rotten oranges in the grid.
• We also initialize a variable to keep track of the total number of fresh oranges.
• We enter a loop while the queue is not empty.
• Inside the loop, we pop an orange from the queue and check its neighboring oranges.
• If a neighboring orange is fresh, we rot it by updating its value to 2 and decrement the count of fresh oranges.
• We enqueue the newly rotten orange and continue the process until the queue is empty.
• Finally, we check if there are any remaining fresh oranges.
• If there are, it means they cannot be rotten, so we return -1. Otherwise, we return the number of minutes.

Explain your approach:

• We will implement a breadth-first search (BFS) algorithm to simulate the rotting process.
• First, we initialize a queue to store the coordinates of the rotten oranges and their respective minutes.
• We iterate over the grid and enqueue the coordinates of the rotten oranges in the queue.
• We also initialize a variable fresh_oranges to keep track of the count of fresh oranges.
• Next, we enter a loop while the queue is not empty.
• Inside the loop, we dequeue an orange from the queue and check its neighboring oranges.
• If a neighboring orange is fresh, we rot it by updating its value to 2 and decrement the fresh_oranges count.
• We enqueue the newly rotten orange and continue the process until the queue is empty.
• After the loop, we check if there are any remaining fresh oranges.
• If there are, it means they cannot be rotten, so we return -1. Otherwise, we return the maximum number of minutes stored in the queue.

Write clean and readable code:

python

class Solution:
    def orangesRotting(self, grid):
        rows = len(grid)
        cols = len(grid[0])

        queue = []
        fresh_oranges = 0

        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 2:
                    queue.append((i, j, 0))
                elif grid[i][j] == 1:
                    fresh_oranges += 1

        minutes = 0

        while queue:
            i, j, minutes = queue.pop(0)

            for di, dj in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                ni, nj = i + di, j + dj

                if 0 <= ni < rows and 0 <= nj < cols and grid[ni][nj] == 1:
                    grid[ni][nj] = 2
                    fresh_oranges -= 1
                    queue.append((ni, nj, minutes + 1))

        return -1 if fresh_oranges != 0 else minutes

solution = Solution()
grid = [[2, 1, 1], [1, 1, 0], [0, 1, 1]]
minutes = solution.orangesRotting(grid)
print(minutes)  # Output: 4

Test your code:

• We test the code with a grid containing both fresh and rotten oranges.
• We chose this test case to ensure that the code can handle a mixed configuration of oranges.
• We expect the output to be 4, which represents the minimum number of minutes required to rot all the fresh oranges.

Optimize if necessary:

• The provided solution already uses the BFS approach, which is the most efficient way to solve this problem.
• There are no further optimizations needed for this solution.

Handle error cases:

• The code assumes that the input grid is a valid 2D list containing only 0, 1, or 2 as values.
• If the input is empty or the grid size is outside the constraints, the behavior of the code may be undefined.

Discuss complexity analysis:

• Let N be the number of cells in the grid.
• The time complexity of the solution is O(N) since each cell is visited at most once during the BFS process.
• The space complexity is O(N) as the queue can store up to N/2 coordinates in the worst case.
• The solution is efficient and performs well even for large grid sizes.