Rotten Oranges
In a grid, each cell can be empty (0), a fresh orange (1), or a rotten orange (2). Every minute, fresh oranges adjacent (4-directionally) to rotten ones become rotten.
Return the minimum minutes until no fresh orange remains, or -1 if impossible.
Input: grid rows separated by newlines, each row comma-separated.
Example:
2,1,1 1,1,0 0,1,1
4
- Initially, the grid is: 2, 1, 1 1, 1, 0 0, 1, 1 with one rotten orange (2) and five fresh oranges (1)
- After the first minute, the fresh oranges adjacent to the rotten one become rotten: 2, 2, 2 1, 1, 0 0, 1, 1
- In the next two minutes, the remaining fresh oranges become rotten: 2, 2, 2 2, 2, 0 0, 2, 2 and then all oranges are rotten after a total of 2+1+1=4 minutes, but since the first minute has already passed, it takes a total of 4 minutes
- The final output is the minimum minutes until no fresh orange remains, which is 4
Constraints:
- 1 <= rows, cols <= 10
- grid[i][j] is 0, 1, or 2
Background Knowledge
The "Rotten Oranges" problem involves a grid data structure, where each cell represents a state (empty, fresh orange, or rotten orange). The problem requires understanding of graph theory, specifically adjacency and neighborhood concepts. In this context, adjacent cells are those that share a common edge (4-directionally: up, down, left, right). The problem also involves time-based simulation, where the state of the grid changes over time based on specific rules.
The key concept here is propagation, where the rotten orange state spreads to adjacent fresh oranges. This can be modeled using graph traversal techniques, such as Breadth-First Search (BFS) or Depth-First Search (DFS). Understanding how to represent the grid as a graph and how to traverse it efficiently is crucial to solving this problem.
The problem also requires consideration of termination conditions, as the simulation needs to stop when no fresh oranges remain or when it's impossible to rot all fresh oranges. This involves understanding base cases and loop termination conditions.
Algorithm/Approach
The general approach to solve this type of problem involves using a graph traversal algorithm to simulate the propagation of the rotten orange state. A suitable algorithm pattern for this problem is Breadth-First Search (BFS), which is particularly useful for solving problems that involve shortest paths or minimum time to reach a certain state.
Step-by-Step Strategy
To implement the solution, follow these steps:
- Represent the input grid as a 2D array or matrix.
- Initialize a queue with the initial rotten oranges (cells with state 2).
- Perform a BFS traversal, where each iteration represents a minute:
- Dequeue a rotten orange and explore its adjacent cells.
- If an adjacent cell is a fresh orange, mark it as rotten and enqueue it.
- Keep track of the number of minutes passed and the number of fresh oranges remaining.
- Continue the BFS traversal until no fresh oranges remain or the queue is empty (indicating that no more fresh oranges can be rotten).
- Return the minimum minutes until no fresh orange remains, or -1 if impossible.
Common Pitfalls
When implementing the solution, watch out for:
- Incorrectly handling the grid boundaries and adjacency checks.
- Failing to update the queue correctly during the BFS traversal.
- Not keeping track of the number of fresh oranges remaining and the number of minutes passed.
- Not handling the termination conditions correctly (e.g., returning -1 when impossible).
Time & Space Complexity
The expected time complexity for this problem is O(m * n), where m and n are the dimensions of the grid, since we need to visit each cell at most once during the BFS traversal. The space complexity is also O(m * n), as we need to store the queue and the grid representation. However, the actual space complexity may be lower if the number of rotten oranges is small, as the queue size will be bounded by the number of rotten oranges.