# 1545. Find Kth Bit in Nth Binary String

## 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)) 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 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 <= 20
• 1 <= k <= 2n - 1

## Solutions

Solution 1: Case Analysis + Recursion

We can observe that for $S_n$, the first half is the same as $S_{n-1}$, and the second half is the reverse and negation of $S_{n-1}$. Therefore, we can design a function $dfs(n, k)$, which represents the $k$-th character of the $n$-th string. The answer is $dfs(n, k)$.

The calculation process of the function $dfs(n, k)$ is as follows:

• If $k = 1$, then the answer is $0$;
• If $k$ is a power of $2$, then the answer is $1$;
• If $k \times 2 < 2^n - 1$, it means that $k$ is in the first half, and the answer is $dfs(n - 1, k)$;
• Otherwise, the answer is $dfs(n - 1, 2^n - k) \oplus 1$, where $\oplus$ represents the XOR operation.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the given $n$ in the problem.

• class Solution {
public char findKthBit(int n, int k) {
return (char) ('0' + dfs(n, k));
}

private int dfs(int n, int k) {
if (k == 1) {
return 0;
}
if ((k & (k - 1)) == 0) {
return 1;
}
int m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
}
}

• class Solution {
public:
char findKthBit(int n, int k) {
function<int(int, int)> dfs = [&](int n, int k) {
if (k == 1) {
return 0;
}
if ((k & (k - 1)) == 0) {
return 1;
}
int m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
};
return '0' + dfs(n, k);
}
};

• class Solution:
def findKthBit(self, n: int, k: int) -> str:
def dfs(n: int, k: int) -> int:
if k == 1:
return 0
if (k & (k - 1)) == 0:
return 1
m = 1 << n
if k * 2 < m - 1:
return dfs(n - 1, k)
return dfs(n - 1, m - k) ^ 1

return str(dfs(n, k))


• func findKthBit(n int, k int) byte {
var dfs func(n, k int) int
dfs = func(n, k int) int {
if k == 1 {
return 0
}
if k&(k-1) == 0 {
return 1
}
m := 1 << n
if k*2 < m-1 {
return dfs(n-1, k)
}
return dfs(n-1, m-k) ^ 1
}
return byte('0' + dfs(n, k))
}

• function findKthBit(n: number, k: number): string {
const dfs = (n: number, k: number): number => {
if (k === 1) {
return 0;
}
if ((k & (k - 1)) === 0) {
return 1;
}
const m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
};
return dfs(n, k).toString();
}