1545.FindKthBitinNthBinaryString.py

"""
    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 4 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:
    Input: n = 3, k = 1
    Output: "0"
    Explanation: S3 is "0111001". The first bit is "0".

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

    Example:
    Input: n = 1, k = 1
    Output: "0"

    Example:
    Input: n = 2, k = 3
    Output: "1"

    Constraints:
        - 1 <= n <= 20
        - 1 <= k <= 2**n - 1
"""
#Difficulty: Medium
#63 / 63 test cases passed.
#Runtime: 612 ms
#Memory Usage: 19.8 MB

#Runtime: 612 ms, faster than 46.50% of Python3 online submissions for Find Kth Bit in Nth Binary String.
#Memory Usage: 19.8 MB, less than 5.60% of Python3 online submissions for Find Kth Bit in Nth Binary String.

class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        s = '0'
        while n > 1:
            n -= 1
            s += '1' + self.reverseS(s)
            if len(s) >= k:
                break
        return s[k-1]

    def reverseS(self, s: str):
        s = list(s[::-1])
        reverse = {'0' : '1', '1' : '0'}
        for i in range(len(s)):
            s[i] = reverse[s[i]]
        return ''.join(s)