# Solving Leetcode Interviews in Seconds with AI: K-th Symbol in Grammar


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```