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

class Solution {
    public char findKthBit(int n, int k) {
        if (n == 1)
            return '0';
        // Calculate length of Sn (2^n - 1)
        int len = (1 << n) - 1;
        int mid = (len / 2) + 1;
        // The middle will always be '1', because of how we create Sn
        if (k == mid)
            return '1';
        else if (k < mid)
            return findKthBit(n - 1, k);
        else
            // Find the answer received from 2nd part, because 2nd part is created
            // by flipping and inverting
            return findKthBit(n - 1, len - k + 1) == '0' ? '1' : '0';
    }
}