Is Snake in Matrix easy or hard?

Snake in Matrix is categorized as an Easy problem. It mainly tests basic grid simulation, coordinate updates, and understanding of row-major indexing. The optimal solution is straightforward and runs in linear time with constant space.

Snake in Matrix Python/Java solution

The implementation is identical across languages. Maintain row and column variables, iterate through the command list, update coordinates for each direction, and compute the final index using row * n + col. This approach works in Python, Java, C++, JavaScript, and C#.

How to solve Snake in Matrix in O(k)?

Initialize row = 0 and col = 0 to represent the starting cell. Iterate through the command list and update the coordinates based on the direction. After processing all moves, compute the final index using row * n + col. This single pass over the commands gives an O(k) time solution.

What is the best approach for Snake in Matrix?

The coordinate mapping simulation is the most common solution. Track the snake’s row and column while processing each direction command, then convert the final position to a linear index using row * n + col. This runs in O(k) time with O(1) space where k is the number of moves.

Is Snake in Matrix asked at Google/Amazon/Meta?

Snake in Matrix is primarily a LeetCode-style simulation problem rather than a frequently reported FAANG interview question. However, similar grid simulation and coordinate tracking problems appear in interviews at companies like Amazon and Google.

What data structure is used in Snake in Matrix?

The solution mainly uses primitive variables to track row and column coordinates. Some implementations also use a small direction map or vector table to convert commands into movement offsets. No complex data structures are required.

What is the time complexity of Snake in Matrix?

The time complexity is O(k), where k is the number of movement commands. Each command results in a constant-time coordinate update. Space complexity remains O(1) since only the current row and column are stored.

Description Hints 1 Solution Video 2:08

3248. Snake in Matrix

Easy81.8% acceptance1 company

There is a snake in an n x n matrix grid and can move in four possible directions. Each cell in the grid is identified by the position: grid[i][j] = (i * n) + j.

The snake starts at cell 0 and follows a sequence of commands.

You are given an integer n representing the size of the grid and an array of strings commands where each command[i] is either "UP", "RIGHT", "DOWN", and "LEFT". It's guaranteed that the snake will remain within the grid boundaries throughout its movement.

Return the position of the final cell where the snake ends up after executing commands.

Example 1:

Input: n = 2, commands = ["RIGHT","DOWN"]

Output: 3

Explanation:

0	1
2	3

0	1
2	3

0	1
2	3

Example 2:

Input: n = 3, commands = ["DOWN","RIGHT","UP"]

Output: 1

Explanation:

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

Constraints:

2 <= n <= 10
1 <= commands.length <= 100
commands consists only of "UP", "RIGHT", "DOWN", and "LEFT".
The input is generated such the snake will not move outside of the boundaries.

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

Try to update the row and column of the snake after each command.

2:08

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
2    def finalPositionOfSnake(self, n: int, commands: List[str]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

2
["RIGHT","DOWN"]

Input

3
["DOWN","RIGHT","UP"]

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