Introduction

In this blog post, we will explore how to solve the LeetCode problem "1545" 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

Given two positive integers n and k, the binary string Sn is formed as follows: S1 = "0" Si = Si - 1 + "1" + reverse(invert(Si - 1)) for i > 1 Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0). For example, the first four strings in the above sequence are: S1 = "0" S2 = "011" S3 = "0111001" S4 = "011100110110001" Return the kth bit in Sn. It is guaranteed that k is valid for the given n. Example 1: Input: n = 3, k = 1 Output: "0" Explanation: S3 is "0111001". The 1st bit is "0". Example 2: Input: n = 4, k = 11 Output: "1" Explanation: S4 is "011100110110001". The 11th bit is "1". Constraints: 1 <= n <= 20 1 <= k <= 2n - 1

Explanation

Here's the approach to solve this problem efficiently:

• Leverage Recursion/Iteration and Symmetry: The core idea is to avoid constructing the entire string Sn. Instead, recursively/iteratively determine which half of the string k falls into, and adjust k and potentially invert the result based on the position and the inversion rule. We can deduce the kth bit based on the properties of how Sn is constructed from Sn-1.
• Exploit the Length of Sn: The length of Sn is 2^n - 1. Use this to decide if the target index 'k' is in the first half, the middle, or the reversed-inverted second half of the string constructed in the previous iteration.

• Early Termination: If k falls right on the middle '1' of the string Sn, then we can directly return '1', avoiding further recursion.

• Runtime Complexity: O(n), where n is the input integer.

• Storage Complexity: O(1). The iterative approach consumes constant extra space.

Code

    def findKthBit(n: int, k: int) -> str:
    """
    Finds the kth bit in the Sn string.

    Args:
        n: The integer n.
        k: The integer k.

    Returns:
        The kth bit in Sn, as a string.
    """

    inverted = False
    while n > 1:
        length = (1 << (n - 1)) - 1  # Length of S(n-1)
        if k == length + 1:
            return "1" if not inverted else "0"
        elif k > length + 1:
            k = 2 * length + 1 - k  # Adjust k for the reversed-inverted part
            inverted = not inverted
        n -= 1

    return "0" if not inverted else "1"