How do you solve Snail Traversal on LeetCode?

Snail Traversal (LeetCode #2624) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the traversal pattern is crucial for filling the matrix correctly.

What is the brute force solution for Snail Traversal?

The brute force approach for Snail Traversal is Brute Force, with O(n) time complexity and O(n) space complexity. The brute-force approach involves creating a 2D array and filling it by iterating through the original array in the snail traversal order. We will alternate between filling columns from top to bottom and bottom to top. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Snail Traversal?

Always validate input before processing to avoid runtime errors. Explain your thought process clearly while coding; it helps interviewers understand your approach. Practice dry-running your code on paper to catch potential errors before implementation.

What are common mistakes when solving Snail Traversal?

Forgetting to check if rowsCount * colsCount equals nums.length, leading to incorrect outputs. Mismanaging indices when filling the matrix, especially when alternating directions.

#2624

Snail Traversal

Medium

ArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

An optimal solution can be achieved by calculating the row and column indices mathematically instead of using nested loops. This reduces the overhead of managing state and simplifies the traversal.

⚙️

Algorithm

3 steps

1Step 1: Validate the input as before.
2Step 2: Create a 2D array for the output.
3Step 3: Use a single loop to fill the matrix by calculating the appropriate row and column indices based on the current index.

solution.py12 lines

1# Full working Python code
2
3def snail(nums, rowsCount, colsCount):
4    if rowsCount * colsCount != len(nums):
5        return []
6    matrix = [[0] * colsCount for _ in range(rowsCount)]
7    for i in range(len(nums)):
8        col = i // rowsCount
9        row = i % rowsCount if col % 2 == 0 else rowsCount - 1 - (i % rowsCount)
10        matrix[row][col] = nums[i]
11    return matrix
12

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through the nums array once, and the space complexity is O(n) for the output matrix.

1Understanding the traversal pattern is crucial for filling the matrix correctly.
2Validating input dimensions prevents unnecessary computations and errors.

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