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
class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        a = [0]
        while k > len(a):
            a.append(1)
            for i in range(len(a) - 2, -1, -1):
                a.append(a[i] ^ 1)
        return str(a[k - 1])
class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        a = [0]
        while k > len(a):
            a.append(1)
            for i in range(len(a) - 2, -1, -1):
                a.append(a[i] ^ 1)
        return str(a[k - 1])