How do you solve Zigzag Grid Traversal With Skip on LeetCode?

Zigzag Grid Traversal With Skip (LeetCode #3417) 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: Zigzag traversal alternates direction with each row.

What is the brute force solution for Zigzag Grid Traversal With Skip?

The brute force approach for Zigzag Grid Traversal With Skip is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves traversing the grid in a zigzag manner while skipping every alternate cell. We simply iterate through each row, alternating the direction of traversal, and collect the values of the cells that we visit. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Zigzag Grid Traversal With Skip?

Zigzag Grid Traversal With Skip uses the following concepts: Array, Matrix, Simulation. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Zigzag Grid Traversal With Skip?

Always clarify the problem statement and constraints before coding. Think about edge cases and how your solution handles them. Explain your thought process as you code to demonstrate your understanding.

What are common mistakes when solving Zigzag Grid Traversal With Skip?

Not handling the edge cases of grid dimensions properly. Confusing the direction of traversal between rows.

#3417

Zigzag Grid Traversal With Skip

Easy

ArrayMatrixSimulationArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages a single traversal of the grid while keeping track of the current position and direction. This reduces unnecessary checks and directly accesses the required cells in a more efficient manner.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty result list.
2Step 2: Loop through each row of the grid.
3Step 3: For each row, calculate the starting index based on the row index (0 for even, 1 for odd).
4Step 4: Use a step of 2 to skip every alternate cell directly.
5Step 5: Append the values to the result list and return it.

solution.py7 lines

1# Full working Python code
2result = []
3for i in range(len(grid)):
4    start = i % 2
5    for j in range(start, len(grid[i]), 2):
6        result.append(grid[i][j])
7print(result)

ℹ

Complexity note: The time complexity is O(n) because we efficiently access only the required cells without unnecessary iterations. The space complexity is O(n) for storing the result list.

1Zigzag traversal alternates direction with each row.
2Skipping cells can be efficiently managed by adjusting the starting index.

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