How do you solve Snake in Matrix on LeetCode?

Snake in Matrix (LeetCode #3248) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m) time complexity and O(1) space complexity. Key insight: Understanding grid movement is crucial.

What is the brute force solution for Snake in Matrix?

The brute force approach for Snake in Matrix is Brute Force, with O(m) time complexity and O(1) space complexity. We can simulate the snake's movement by tracking its position in the grid. For each command, we update the position based on the direction specified. This approach is straightforward and easy to understand. The optimal Optimal Solution approach improves this to O(m).

What algorithm or data structure is used to solve Snake in Matrix?

Snake in Matrix uses the following concepts: Array, String, Simulation. The recommended patterns to study are: Simulation, Counting.

What are the interview tips for Snake in Matrix?

Clarify the problem statement and constraints. Think aloud to show your thought process. Test your solution with edge cases.

What are common mistakes when solving Snake in Matrix?

Not considering the grid boundaries. Confusing row and column updates.

#3248

Snake in Matrix

Easy

ArrayStringSimulationSimulationCounting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m)	O(m)
Space	O(1)	O(1)

💡

Intuition

Time O(m)Space O(1)

The optimal solution is similar to the brute force approach but focuses on minimizing unnecessary operations. We can directly calculate the final position based on the number of moves in each direction instead of updating the position step-by-step.

⚙️

Algorithm

4 steps

1Step 1: Initialize counters for row and column movements.
2Step 2: Count the number of moves in each direction (UP, DOWN, LEFT, RIGHT).
3Step 3: Calculate the final row and column based on the counts.
4Step 4: Return the final position using the formula (final_row * n) + final_col.

solution.py6 lines

1def snake_position(n, commands):
2    row_moves = commands.count('DOWN') - commands.count('UP')
3    col_moves = commands.count('RIGHT') - commands.count('LEFT')
4    final_row = row_moves
5    final_col = col_moves
6    return final_row * n + final_col

ℹ

Complexity note: The time complexity remains O(m) since we still iterate through the commands. The space complexity is O(1) as we only use a few counters.

1Understanding grid movement is crucial.
2Counting moves can simplify the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.