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 > 1Where + 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 = "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 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 <= 201 <= k <= 2n - 1class Solution {
public char findKthBit(int n, int k) {
StringBuilder sb = new StringBuilder("0");
for (int i = 1; i < n; i++) {
char[] cs = sb.reverse().toString().toCharArray();
for (int j = 0; j < cs.length; j++)
cs[j] = cs[j] == '0' ? '1' : '0';
sb.reverse();
sb.append("1" + new String(cs));
// System.out.println(sb.toString());
}
return sb.charAt(k - 1);
}
}