# 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

Solution:

class Solution {
public char findKthBit(int n, int k) {
/*
S1 = "0"                     1
S2 = "011"                   3
S3 = "0111001"               7
S4 = "0111001 1 0110001"       15
*/
// System.out.println(n + ", " + k);
if (n == 1 && k == 1) return '0';
if (n == 2 && k == 2) return '1';
if (n == 2 && k == 3) return '1';
int len = (int) (Math.pow(2, n)) - 1;
if (k == len / 2 + 1) return '1';
if (k <= len / 2) return findKthBit(n - 1, k);
int nk = len - k + 1;
return findKthBit(n - 1, nk) - '0' == 0 ? '1' : '0';
}
}