Find Kth Bit in Nth Binary String

Problem

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 = "0**1**1"
S3 = "011**1**001"
S4 = "0111001**1**0110001"

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

Solution (Java)

class Solution {
    public char findKthBit(int n, int k) {
        boolean flip = false;
        while (k != 1) {
            int base = floorTwo(k);
            if (base == k) {
                return flip ? '0' : '1';
            }
            flip = !flip;
            k = base - (k - base);
        }
        return flip ? '1' : '0';
    }

    private int floorTwo(int k) {
        while ((k & (k - 1)) > 0) {
            k &= k - 1;
        }
        return k;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).