Introduction

In this blog post, we will explore how to solve the LeetCode problem "779" 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

We build a table of n rows (1-indexed). We start by writing 0 in the 1st row. Now in every subsequent row, we look at the previous row and replace each occurrence of 0 with 01, and each occurrence of 1 with 10. For example, for n = 3, the 1st row is 0, the 2nd row is 01, and the 3rd row is 0110. Given two integer n and k, return the kth (1-indexed) symbol in the nth row of a table of n rows. Example 1: Input: n = 1, k = 1 Output: 0 Explanation: row 1: 0 Example 2: Input: n = 2, k = 1 Output: 0 Explanation: row 1: 0 row 2: 01 Example 3: Input: n = 2, k = 2 Output: 1 Explanation: row 1: 0 row 2: 01 Constraints: 1 <= n <= 30 1 <= k <= 2n - 1

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Leverage Symmetry/Recursion: The core idea is to recognize the recursive/self-similar nature of the row generation. Each row is built by transforming the previous row. We can recursively determine the k-th element by relating it to its "parent" element in the previous row.
• Focus on Position Transformation: Instead of generating the entire row, we track how the position k transforms as we move up the rows. If k is in the second half of the row, it originates from transforming an element in the previous row, effectively making the problem smaller.

• Parity Determines Result: The initial value is 0. Whether a transformation flips the value (0 to 1, or 1 to 0) depends on whether the number of transformations undergone on that element is even or odd. We can efficiently determine this without explicitly simulating the transformations.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def kth_grammar(n: int, k: int) -> int:
    """
    Finds the kth symbol in the nth row of the grammar table.

    Args:
        n: The row number (1-indexed).
        k: The symbol position in the row (1-indexed).

    Returns:
        The kth symbol (0 or 1).
    """

    result = 0
    while n > 1:
        length = 2**(n - 2)  # Length of half of the previous row
        if k > length:
            result = 1 - result  # Flip the result due to transformation
            k -= length
        n -= 1

    return result