Description
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))fori > 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 = "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 - 1
Solutions
This function finds the kth character in the nth iteration of a recursive binary sequence. It maintains two complementary sequences stored in arrays s and r: each new level of s is built by concatenating the previous s, a pivot bit '1', and the previous r (which represents the bitwise complement of s). Similarly, r is built from the previous s, a pivot '0', and the previous r. The loop runs until the current sequence s is long enough to contain the kth position, then returns the character at index k-1 (converting 1-indexed input to 0-indexed array access). For example, with n=1, s[0]="0"; with n=2, s[1]="011"; with n=3, s[2]="0111001", and so on.
/**
* @param {number} n
* @param {number} k
* @return {character}
*/
var findKthBit = function(n, k) {
let s = ['0'];
let r = ['1'];
let pos = 1;
while (s[pos - 1].length < k) {
s[pos] = s[pos - 1] + '1' + r[pos - 1];
r[pos] = s[pos - 1] + '0' + r[pos - 1];
pos++;
}
return s[pos - 1][k - 1];
};