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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.

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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))

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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;
    }
}

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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);
    }
};

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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))
}

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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();
}

Solution 2: Bit Manipulation

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const findKthBit = (n: number, k: number): string =>
    String((((k / (k & -k)) >> 1) & 1) ^ (k & 1) ^ 1);

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const findKthBit = (n, k) => String((((k / (k & -k)) >> 1) & 1) ^ (k & 1) ^ 1);

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