What is the optimal time complexity of solving Rotting Oranges?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Rotting Oranges?

The space complexity is O(1) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Rotting Oranges?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Rotting Oranges?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Rotting Oranges

Learn how to solve the 'Rotting Oranges' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

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.

Write a function orangesRotting(grid: List[List[int]]) -> int.

Constraints

•m == len(grid)
•n == len(grid[i])
•1 <= m, n <= 10
•grid[i][j] is 0, 1, or 2

Examples

Example 1

Input

grid = [[2,1,1],[1,1,0],[0,1,1]]

Output

Explanation

Minute 0: rotten at (0,0). Fresh at (0,1), (0,2), (1,0), (1,1), (2,1), (2,2). Minute 1: fresh at (0,1) and (1,0) rot. Minute 2: fresh at (0,2) and (1,1) rot. Minute 3: fresh at (2,1) rot. Minute 4: fresh at (2,2) rot.

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 adjacent to a rotten orange, so it stays fresh.

Need a Hint?

Represent graph node connections as an adjacency list/matrix, then use standard BFS or DFS graph traversal.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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