1545. Find Kth Bit in Nth Binary String

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

Approach 01:

#include <cmath>

class Solution {
public:
    char findKthBit(int n, int k) {
        if (n == 1)
            return '0';
        
        const int midIndex = pow(2, n - 1); // 1-indexed
        
        if (k == midIndex)
            return '1';
        if (k < midIndex)
            return findKthBit(n - 1, k);
        
        return findKthBit(n - 1, midIndex * 2 - k) == '0' ? '1' : '0';
    }
};

class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        if n == 1:
            return '0'

        midIndex = 2 ** (n - 1)  # 1-indexed

        if k == midIndex:
            return '1'
        if k < midIndex:
            return self.findKthBit(n - 1, k)

        return '1' if self.findKthBit(n - 1, midIndex * 2 - k) == '0' else '0'

Time Complexity

• Recursive Calls:

  The function makes recursive calls based on the value of n. Specifically, each call reduces n by 1 until it reaches 1. Therefore, the number of recursive calls is proportional to n.

• Overall Time Complexity:

  Thus, the overall time complexity is \(O(n)\).

Space Complexity

• Call Stack:

  The maximum depth of the recursion stack is n, as the function can be called n times before reaching the base case. Therefore, the space complexity due to the recursion call stack is \(O(n)\).

• Overall Space Complexity:

  The overall space complexity is \(O(n)\).