1545. Find Kth Bit in Nth Binary String

Problem Link

Description

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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

Solution

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Python3

class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        
        @cache
        def go(x):
            if x <= 1: return "0"
            
            last = go(x - 1)
            
            r = "".join(["1" if c == "0" else "0" for c in last])
            
            return go(x - 1) + "1" + r[::-1]
        
        return go(n)[k - 1]