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 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 <= 201 <= k <= 2n - 1
Related Topics:
String
The length of the string len is 2^n - 1.
If k - 1 == len / 2, then this is the middle of the string, return 1 unless n == 1.
If k - 1 < len / 2, this is at the left part of Sn, which is the same as findKthBit(n - 1, k).
If k - 1 > len / 2, this is the i = k - 1 - len / 2-th bit in the right part, which is the invert of findKthBit(n - 1, len / 2 - i + 1).
// OJ: https://leetcode.com/problems/find-kth-bit-in-nth-binary-string/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
char findKthBit(int n, int k) {
if (n == 1) return '0';
int len = pow(2, n) - 1;
if (k - 1 == len / 2) return '1';
if (k - 1 < len / 2) return findKthBit(n - 1, k);
int i = k - 1 - len / 2;
return findKthBit(n - 1, len / 2 - i + 1) == '0' ? '1' : '0';
}
};