Find Kth Bit in Nth Binary String

LeetCode 1545, Medium. Topics: String, Recursion, Simulation. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 18 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1545   # by problem number
lcpy gen -s find_kth_bit_in_nth_binary_string   # by problem name

Problem

Given two positive integers n and k, the binary string S_n is formed as follows:

S_1 = "0"
S_i = S_i - 1 + "1" + reverse(invert(S_i - 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:

S_1 = "0"
S_2 = "011"
S_3 = "0111001"
S_4 = "011100110110001"

Return the k^th bit in S_n. It is guaranteed that k is valid for the given n.

Examples

Input: n = 3, k = 1
Output: "0"
Explanation: S3 is "0111001".
The 1st bit is "0".

Input: n = 4, k = 11
Output: "1"
Explanation: S4 is "011100110110001".
The 11th bit is "1".

Constraints

1 <= n <= 20
1 <= k <= 2^n - 1

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(log k)
    # Space: O(log k)
    def find_kth_bit(self, n: int, k: int) -> str:
        if k == 1:
            return "0"
        half = 1
        while half * 2 + 1 < k:
            half = half * 2 + 1
        mid = half + 1
        if k == mid:
            return "1"
        mirrored = self.find_kth_bit(n, mid - (k - mid))
        return "0" if mirrored == "1" else "1"

Complexity

Time	Space
O(log k)	O(log k)

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Find Kth Bit in Nth Binary String

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