human
I found an interesting problem on https://leetcode.com/problems/find-kth-bit-in-nth-binary-string/:
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 4 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 first bit is "0".

Example 2:
Input: n = 4, k = 11
Output: "1"
Explanation: S4 is "011100110110001". The 11th bit is "1".

Example 3:
Input: n = 1, k = 1
Output: "0"

Example 4:
Input: n = 2, k = 3
Output: "1"

 
Constraints:

1 <= n <= 20
1 <= k <= 2n - 1

I tried it in Python, but could not do it. Can you solve it?

 ... Can you provide a the Python program you requested?
assistant
#Here is the Python program you requested.

class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        def reverse_invert(array: List[str]) -> List[str]:
            res = []
            for ind in range(len(array)-1, -1, -1):
                res.append('1' if array[ind] == '0' else '0')
            return res
        
        bits = ['0']
        for curr in range(2, n+1):
            bits += ['1'] + reverse_invert(bits)
        return bits[k-1]