Introduction

In this blog post, we will explore how to solve the LeetCode problem "3417" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an m x n 2D array grid of positive integers. Your task is to traverse grid in a zigzag pattern while skipping every alternate cell. Zigzag pattern traversal is defined as following the below actions: Start at the top-left cell (0, 0). Move right within a row until the end of the row is reached. Drop down to the next row, then traverse left until the beginning of the row is reached. Continue alternating between right and left traversal until every row has been traversed. Note that you must skip every alternate cell during the traversal. Return an array of integers result containing, in order, the value of the cells visited during the zigzag traversal with skips. Example 1: Input: grid = [[1,2],[3,4]] Output: [1,4] Explanation: Example 2: Input: grid = [[2,1],[2,1],[2,1]] Output: [2,1,2] Explanation: Example 3: Input: grid = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,3,5,7,9] Explanation: Constraints: 2 <= n == grid.length <= 50 2 <= m == grid[i].length <= 50 1 <= grid[i][j] <= 2500

Explanation

• Traverse the grid row by row, alternating the direction of traversal in each row.
  • In each row, visit only the cells at even indices if moving right, and even indices from the right if moving left.
  • Collect the visited cell values into the result array.

• Runtime Complexity: O(m*n), Storage Complexity: O(m*n)

Code

    def zigzag_skip(grid):
    """
    Traverses a 2D grid in a zigzag pattern, skipping alternate cells.

    Args:
        grid: A list of lists of integers representing the 2D grid.

    Returns:
        A list of integers representing the values of the visited cells.
    """

    result = []
    m = len(grid)
    n = len(grid[0])
    direction = 1  # 1 for right, -1 for left

    for i in range(m):
        if direction == 1:
            for j in range(0, n, 2):
                result.append(grid[i][j])
        else:
            for j in range(n - 1 if n % 2 != 0 else n - 2, -1, -2):
                result.append(grid[i][j])
        direction *= -1
    return result