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 <= 201 <= k <= 2n - 1
Solutions
Solution 1: Case Analysis + Recursion
We can observe that for Sn, the first half is the same as Sn-1, and the second half is the reverse and negation of Sn-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 × 2 < 2n - 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, 2n - 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.
PythonJavaC++GoTypeScriptRustJavaScript
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))(code-box)
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;
}
}(code-box)
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);
}
};(code-box)
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))
}(code-box)
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();
}(code-box)
impl Solution {
pub fn find_kth_bit(n: i32, k: i32) -> char {
fn dfs(n: i32, k: i32) -> i32 {
if k == 1 {
return 0;
}
if (k & (k - 1)) == 0 {
return 1;
}
let m: i32 = 1 << n;
if k * 2 < m - 1 {
dfs(n - 1, k)
} else {
dfs(n - 1, m - k) ^ 1
}
}
if dfs(n, k) == 0 { '0' } else { '1' }
}
}(code-box)
/**
* @param {number} n
* @param {number} k
* @return {character}
*/
var findKthBit = function (n, k) {
const dfs = function (n, k) {
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();
};(code-box)
