In the world of programming challenges, navigating through grids and matrices is a common task that tests your problem-solving skills. One intriguing problem that often appears in coding interviews is tracking the movement of a snake within an n x n
matrix. If you’re keen on improving your coding skills or simply enjoy tackling interesting challenges, understanding how to solve this problem can be both educational and fun.
Problem Question | Snake in Matrix | Leetcode 3248
Imagine a matrix of size n x n
, where each cell is uniquely identified by a number derived from its row and column indices. For instance, the cell at position (i, j)
is represented by the formula i * n + j
. A snake starts at the top-left cell (position 0) and moves based on a series of commands. These commands can be "UP", "DOWN", "LEFT", and "RIGHT". Your goal is to determine the final position of the snake after executing all the commands.
Example Breakdown
Consider a 2x2 matrix and a sequence of commands like ["RIGHT", "DOWN"]
. Here’s how the snake’s position changes:
- Initial Position: The snake starts at cell 0.
- Command: RIGHT: Moves from cell 0 to cell 1.
- Command: DOWN: Moves from cell 1 to cell 3.
So, the final position is cell 3. Simple, right? Now, let’s delve into the solution approach that ensures accuracy and efficiency.
Step-by-Step Solution Approach
To solve the problem of tracking the snake's movement in an n x n
grid, follow these steps:
1. Initialize the Starting Point
Start by setting the initial position of the snake at the top-left corner of the matrix. In terms of indices, this is (i = 0, j = 0)
where i
represents the row index and j
represents the column index.
2. Process Each Command
Iterate through the list of commands and adjust the position of the snake according to the direction specified:
- "RIGHT": Increment the column index
j
. - "DOWN": Increment the row index
i
. - "UP": Decrement the row index
i
, but only ifi > 0
to avoid going out of bounds. - "LEFT": Decrement the column index
j
, but only ifj > 0
to avoid going out of bounds.
3. Compute the Final Position
Once all commands are processed, compute the final position of the snake using the formula (i * n) + j
. This converts the row and column indices into a single cell index based on the matrix size.
Code Explanation
- Initialization: The variables
i
andj
are initialized to 0, representing the starting cell. - Command Processing: For each command, the corresponding index is updated. Boundary conditions are checked to ensure the snake doesn’t move out of bounds.
- Final Position Calculation: The final position is computed using the formula
(i * n) + j
, which provides the cell number in a linearized matrix.
Key Takeaways
- Efficiency: The algorithm processes each command in O(m) time, where
m
is the number of commands. This ensures that it runs efficiently even with the maximum constraints. - Boundary Handling: The checks for boundaries (i.e.,
i > 0
for "UP" andj > 0
for "LEFT") prevent the snake from moving out of the grid, making the solution robust and reliable.
n x n
matrix is a fantastic way to practice grid navigation and boundary handling in programming. By following a clear and methodical approach, you can efficiently solve this problem and gain valuable insights into matrix manipulations. Whether you're preparing for coding interviews or just honing your problem-solving skills, mastering such challenges can significantly enhance your programming prowess.