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

Example 2:
Input: n = 4, k = 11
Output: "1"
Explanation: S4 is "011100110110001". The 11th bit is "1".

Example 3:
Input: n = 1, k = 1
Output: "0"

Example 4:
Input: n = 2, k = 3
Output: "1"

 
Constraints:

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

Difficulty: MEDIUM
Tags: String algorithms

Complete search
Skill types: Complete search

Starter code:
class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        

Solutions:
class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		i = n - 1
		invert = False
		while i > 0:
			half_len = (2 ** (i + 1) - 1) // 2
			if k == half_len + 1:
				return '1' if not invert else '0'
			if k > half_len:
				k = half_len - (k - half_len - 1) + 1
				invert = not invert
			i -= 1
		return '1' if invert else '0'


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		return str(k // (k & -k) >> 1 & 1 ^ (k & 1 ^ 1))

	def findKthBit_logn(self, n: int, k: int) -> str:
		(cur, flip) = (1 << n - 1, 0)
		while k > 1:
			if k & k - 1 == 0:
				return str(flip ^ 1)
			if k > cur:
				k = 2 * cur - k
				flip ^= 1
			cur >>= 1
		return str(flip)


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		dp = [[]] * n
		dp[0] = [0]
		for i in range(1, n):
			dp[i] = dp[i - 1] + [1] + list(reversed([v ^ 1 for v in dp[i - 1]]))
		return str(dp[-1][k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(s):
			return ''.join(['1' if c == '0' else '0' for c in s])

		def revStr(s):
			return ''.join(list(reversed(s)))
		counter = 1
		x = '0'
		while counter != n:
			x = x + '1' + revStr(invert(x))
			counter += 1
		return x[k - 1]


def reverse(m):
	t = ''
	for i in m:
		if i == '1':
			t = t + '0'
		else:
			t = t + '1'
	return t

class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = '0'
		while n > 1:
			t = reverse(s)
			s = s + '1' + t[::-1]
			n -= 1
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		return self.solv(n, k)

	def solv(sef, n, k):
		i = 1
		lst = ['0']
		ans = '0'
		ki = 1
		if k == 1:
			return '0'
		while i < n + 1:
			ans = '1'
			lst.append(ans)
			ki += 1
			if ki == k:
				return ans
			for j in range(ki - 2, -1, -1):
				ans = '1' if lst[j] == '0' else '0'
				lst.append(ans)
				ki += 1
				if ki == k:
					return ans
			i += 1
		return None


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		full = [0]
		for _ in range(n - 1):
			full = full + [1] + [(x + 1) % 2 for x in full][::-1]
		return str(full[k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		l = [0] * (n + 1)
		for i in range(1, n + 1):
			l[i] = 2 * l[i - 1] + 1

		def helper(n, k, f):
			m = l[n]
			if n == 1:
				return 0 ^ f
			if k == m // 2 + 1:
				return 1 ^ f
			if k <= m // 2:
				return helper(n - 1, k, f)
			else:
				return helper(n - 1, m - k + 1, 1 ^ f)
		return str(helper(n, k, 0))


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		S = [0]
		for i in range(2, n + 1):
			S += [1] + [1 if i == 0 else 0 for i in S[::-1]]
		return str(S[k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		cache = [''] * n

		def gen_str(n):
			if n - 1 in cache:
				return cache[n - 1]
			if n == 1:
				s = '0'
				cache[0] = s
				return s
			prev = gen_str(n - 1)
			s = prev + '1'
			for c in reversed(prev):
				if c == '0':
					s += '1'
				else:
					s += '0'
			cache[n - 1] = s
			return s
		return gen_str(n)[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(data):
			ans = ''
			for val in data:
				if val == '1':
					ans += '0'
				else:
					ans += '1'
			return ans
		res = [''] * (n + 1)
		res[1] = '0'
		for i in range(2, n + 1):
			res[i] = res[i - 1] + '1' + invert(res[i - 1])[::-1]
		return res[n][k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = '0'
		while n > 1:
			p = ''
			for l in s[::-1]:
				p += '1' if l == '0' else '0'
			s = s + '1' + p
			n -= 1
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		dp = [None] * n
		dp[0] = '0'
		for i in range(1, n):
			m = ''
			inv = self.invert(dp[i - 1], m)
			rev = self.reverse(inv)
			dp[i] = dp[i - 1] + '1' + rev
		return dp[-1][k - 1]

	def reverse(self, string):
		return string[::-1]

	def invert(self, string, m):
		for i in range(len(string)):
			if string[i] == '1':
				m += '0'
			else:
				m += '1'
		return m


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(s):
			inv = ''
			for ch in s:
				inv += '0' if ch == '1' else '1'
			return inv
		s = '0'
		for i in range(1, n):
			s = s + '1' + invert(s)[::-1]
		return s[k - 1]


class Solution:

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

		def rev(s):
			return s[::-1]

		def invert(r):
			t = ''
			for i in r:
				if i == '1':
					t += '0'
				else:
					t += '1'
			return t
		z = '0'
		for i in range(2, n + 1):
			a = invert(z)
			b = rev(a)
			z += '1' + b
		return z[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = []
		s.append('0')
		for i in range(1, n):
			ans = s[i - 1] + '1'
			temp = ''
			for j in s[i - 1]:
				if j == '0':
					temp += '1'
				else:
					temp += '0'
			ans = ans + temp[::-1]
			s.append(ans)
		return s[-1][k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def reverse(cur_s):
			return ''.join([c for c in reversed(cur_s)])

		def invert(cur_s):
			return ''.join(['1' if c == '0' else '0' for c in cur_s])
		S = '0'
		for i in range(1, n):
			S += '1' + reverse(invert(S))
		return S[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		x = '0'
		for i in range(1, n):
			x = x + '1' + ''.join(list(map(lambda x: '1' if x == '0' else '0', x))[::-1])
		return x[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		series = ['0']
		for i in range(n - 1):
			series = self.getNext(series)
		return series[k - 1]

	def getNext(self, series):
		extension = series[::-1]
		for (idx, c) in enumerate(extension):
			if c == '0':
				extension[idx] = '1'
			else:
				extension[idx] = '0'
		return series + ['1'] + extension


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = '0'
		for i in range(1, n):
			chars = []
			for j in reversed(list(range(len(s)))):
				chars.append('1' if s[j] == '0' else '0')
			s += '1' + ''.join(chars)
		return s[k - 1]


def invert(s):
	return bin(int(s, 2) ^ int('1' * len(s), 2))[2:]

def reverse(s):
	return s[::-1]

class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		S = [None] * n
		S[0] = '0'
		for i in range(1, n):
			S[i] = S[i - 1] + '1' + reverse(invert(S[i - 1]))
		return S[n - 1][k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		bits = [False]
		for i in range(1, n + 1):
			bits.extend([True] + [j == False for j in bits[::-1]])
		return str(int(bits[k - 1]))


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(s):
			temp = ''
			for i in range(len(s)):
				if s[i] == '1':
					temp = temp + '0'
				else:
					temp = temp + '1'
			return temp
		s = '0'
		for i in range(1, n):
			inv = invert(s)
			rev = inv[::-1]
			s = s + '1' + rev
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = ['0']
		for i in range(n - 1):
			temp1 = s.copy()
			for i in range(len(temp1)):
				if temp1[i] == '0':
					temp1[i] = '1'
				elif temp1[i] == '1':
					temp1[i] = '0'
			temp1.reverse()
			temp = s + ['1'] + temp1
			s = temp
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def reverse_invert(array: List[str]) -> List[str]:
			res = []
			for ind in range(len(array) - 1, -1, -1):
				res.append('1' if array[ind] == '0' else '0')
			return res
		bits = ['0']
		for curr in range(2, n + 1):
			bits += ['1'] + reverse_invert(bits)
		return bits[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def negVal(val):
			lenVal = len(val)
			mask = int('1' * lenVal, 2)
			val = int(val, 2)
			neg = val ^ mask
			lenNeg = len(bin(neg)) - 2
			neg = '0' * (lenVal - lenNeg) + bin(neg)[2:]
			return neg[::-1]
		strVal = '0'
		for i in range(n - 1):
			strVal += '1' + negVal(strVal)
		if k > len(strVal):
			return '0'
		else:
			return strVal[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(bin):
			bin = list(bin)
			for i in range(len(bin)):
				if bin[i] == '1':
					bin[i] = '0'
				else:
					bin[i] = '1'
			return ''.join(bin)
		bin_n = bin(n)[2:]
		dp = ['0'] * n
		for i in range(1, n):
			dp[i] = dp[i - 1] + '1' + invert(dp[i - 1])[::-1]
		return dp[-1][k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		d = dict()
		d[0] = [0]
		for i in range(1, n + 1):
			d[i] = d[i - 1] + [1] + [1 - x for x in d[i - 1]][::-1]
		return str(d[n][k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		sn = ['0']
		for i in range(1, n):
			nx = sn[i - 1] + '1' + self.reverse(self.invert(sn[i - 1]))
			sn.append(nx)
		return sn[-1][k - 1]

	def invert(self, s):
		return ''.join(('0' if _ == '1' else '1' for _ in s))

	def reverse(self, s):
		return ''.join(reversed(s))


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(n):
			s = ''
			for char in n:
				if char == '0':
					s += '1'
				else:
					s += '0'
			return s

		def generater(n):
			dp = [-1] * n
			dp[0] = '0'
			for i in range(1, len(dp)):
				dp[i] = dp[i - 1] + '1' + invert(dp[i - 1])[::-1]
			return dp[-1]
		return generater(n)[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = ['0']
		for _ in range(n - 1):
			s += ['1'] + [('1', '0')[int(x)] for x in reversed(s)]
		return s[k - 1]


import math

class Solution:

	def findKthBit(self, j: int, k: int) -> str:
		n = j
		lst = ['0']
		for i in range(n):
			lst = lst + ['1'] + ['1' if ele == '0' else '0' for ele in lst[::-1]]
		return lst[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		k -= 1
		s = '0'
		i = 0
		while i < n:
			inverted = ''.join(('1' if c == '0' else '0' for c in s))[::-1]
			s = s + '1' + inverted
			i += 1
		return s[k]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		memo_table = [''] * n
		memo_table[0] = '0'
		for i in range(1, n):
			memo_table[i] = self.new_level(memo_table[i - 1])
		return memo_table[n - 1][k - 1]

	def new_level(self, old_level: str):
		new_level = ['1'] * (len(old_level) * 2 + 1)
		left = 0
		right = len(old_level) * 2
		for c in old_level:
			new_level[left] = c
			if c == '0':
				new_level[right] = '1'
			else:
				new_level[right] = '0'
			left += 1
			right -= 1
		return ''.join(new_level)


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def revert(s):
			ans = ''
			for x in s:
				if x == '1':
					ans += '0'
				else:
					ans += '1'
			return ans

		def helper(n):
			if n == 1:
				return '0'
			res = helper(n - 1)
			return res + '1' + revert(res)[::-1]
		return helper(n)[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		from copy import deepcopy

		@lru_cache()
		def help(i):
			if i == 1:
				return '0'
			s = help(i - 1)
			s2 = deepcopy(s)
			v = []
			for i in s:
				v.append(i)
			for i in range(len(v)):
				if v[i] == '1':
					v[i] = '0'
				else:
					v[i] = '1'
			v = v[::-1]
			s = ''.join(v)
			return s2 + '1' + s
		return help(n)[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def reverseInvert(string: str) -> str:
			output = []
			for char in string:
				if char == '0':
					output.append('1')
				else:
					output.append('0')
			output.reverse()
			return ''.join(output)

		def getNthSequence(n: int) -> str:
			if n == 1:
				return '0'
			prev = getNthSequence(n - 1)
			return prev + '1' + reverseInvert(prev)
		return getNthSequence(int(ceil(log2(k)) + 1))[k - 1]


class Solution:

	def invert(self, s):
		out = ''
		for c in s:
			if c == '0':
				out += '1'
			if c == '1':
				out += '0'
		return out

	def reverse(self, s):
		return s[::-1]

	def findKthBit(self, n: int, k: int) -> str:
		l = []
		for i in range(n):
			if i == 0:
				l.append('0')
			else:
				last_s = l[-1]
				s = last_s + '1' + self.reverse(self.invert(last_s))
				l.append(s)
		s = l[-1]
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def inverse(s):
			res = ''
			for ch in s:
				if ch == '1':
					res += '0'
				else:
					res += '1'
			return res
		s = '0'
		for i in range(n - 1):
			s = s + '1' + inverse(s)[::-1]
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s0 = '0'
		i = 1
		while i <= n:
			s0 = s0 + '1' + self.reverse(s0)
			i = i + 1
		return s0[k - 1]

	def reverse(self, s):
		mapp = {'1': '0', '0': '1'}
		ans = ''
		for x in s:
			ans = ans + mapp[x]
		return ans[::-1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def reverse(s):
			return ''.join([s[len(s) - 1 - i] for i in range(len(s))])

		def invert(s):
			return ''.join(['1' if s[i] == '0' else '0' for i in range(len(s))])

		def findNthStr(n1):
			if n1 == 1:
				return '0'
			else:
				prev = findNthStr(n1 - 1)
				return prev + '1' + reverse(invert(prev))
		return findNthStr(n)[k - 1]


class Solution:

	def invert(self, t):
		ret = ''
		for i in t:
			if i == '0':
				ret += '1'
			else:
				ret += '0'
		return ret

	def findKthBit(self, n: int, k: int) -> str:
		i = 1
		l = list()
		l.append('0')
		while i <= n:
			t = l[-1]
			ans = t + '1'
			ans += self.invert(t[::-1])
			l.append(ans)
			i += 1
		return l[-1][k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		dp = ['0'] * n
		for i in range(1, n):
			dp[i] = dp[i - 1] + '1' + ''.join([str(1 - int(n)) for n in dp[i - 1]])[::-1]
		return dp[-1][k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = '0'
		inv = '1'
		rev_inv = '1'
		for i in range(n - 1):
			s = s + '1' + rev_inv
			inv = ''
			rev_inv = ''
			for j in range(len(s)):
				if s[j] == '0':
					inv += '1'
				else:
					inv += '0'
			rev_inv = inv[::-1]
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = []
		for i in range(n):
			if i == 0:
				s.append(['0'])
			else:
				temp = s[i - 1].copy()
				for j in range(len(temp)):
					if temp[j] == '0':
						temp[j] = '1'
					else:
						temp[j] = '0'
				temp = temp[::-1]
				s.append([])
				for zz in s[i - 1]:
					s[i].append(zz)
				s[i].append('1')
				for zz in temp:
					s[i].append(zz)
		return s[-1][k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def helper(memo):
			ans = ''
			for c in memo:
				if c == '0':
					ans += '1'
				else:
					ans += '0'
			return ans
		memo = '0'
		for i in range(1, n + 1):
			memo = memo + '1' + helper(memo)[::-1]
		return memo[k - 1]


bins = [[0]]
for _ in range(20):
	bins.append(bins[-1] + [1] + [1 - x for x in reversed(bins[-1])])

class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		return str(bins[n - 1][k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		if n == 1:
			return '0'
		if n == 2:
			return ['0', '1', '1'][k - 1]
		record = [0, 1, 1]
		for i in range(n - 2):
			record = record + [1] + record
			l = len(record)
			record[l // 2 + l // 2 // 2 + 1] = 0
		return str(record[k - 1])


class Solution:

	def invert(s):
		inv_s = ''
		for char in s:
			if char == '1':
				inv_s += '0'
			else:
				inv_s += '1'
		return inv_s
	answer = ['0']
	for i in range(1, 21):
		prev_s = answer[i - 1]
		answer.append(prev_s + '1' + invert(prev_s)[::-1])

	def findKthBit(self, n: int, k: int) -> str:
		return self.answer[n - 1][k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = '0'
		for i in range(2, n + 1):
			sRevInv = ''.join(['1' if c == '0' else '0' for c in s[::-1]])
			s = s + '1' + sRevInv
			if len(s) >= k:
				return s[k - 1]
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(x):
			return [0 if i == 1 else 1 for i in x]
		S = [0]
		i = 1
		while i < n and len(S) < k:
			S.append(1)
			S.extend(invert(S[:-1])[::-1])
			i += 1
		return str(S[k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def operator(s):
			t = reversed(['0' if i == '1' else '1' for i in s])
			return s + '1' + ''.join(t)
		s = '0'
		while len(s) < k:
			s = operator(s)
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		switch = {'0': '1', '1': '0'}
		s = '0'
		while len(s) < k:
			s = s + '1' + ''.join((switch[i] for i in reversed(s)))
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		l = [0]
		for _ in range(n - 1):
			l = l + [1] + [1 - b for b in reversed(l)]
			if len(l) >= k:
				return str(l[k - 1])
		return str(l[k - 1])


class Solution:
	s = [0]

	def findKthBit(self, n: int, k: int) -> str:
		while len(self.s) < k:
			self.s = self.s + [1] + [n ^ 1 for n in self.s[::-1]]
		return str(self.s[k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def inner(n, k):
			if n == 1:
				re = 0
			elif k == 2 ** (n - 1):
				re = 1
			elif k < 2 ** (n - 1):
				re = inner(n - 1, k)
			else:
				re = inner(n - 1, 2 ** n - k) ^ 1
			return re
		return str(inner(n, k))


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(l):
			result = []
			for item in l:
				if item == '0':
					result.append('1')
				else:
					result.append('0')
			return result
		string = '0'
		while len(string) < k:
			string = string + '1'
			addition = invert(string[0:len(string) - 1])
			addition.reverse()
			addition = ''.join(addition)
			string = string + addition
		return string[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = ''

		def recurse(n):
			nonlocal s
			if n <= 1:
				return '0'
			else:
				_str = recurse(n - 1)
				s = _str + '1' + self.reverse(self.invert(_str))
				return s
		return recurse(n)[k - 1]

	def invert(self, string):
		return ''.join(['0' if char == '1' else '1' for char in string])

	def reverse(self, string):
		return string[::-1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		cur = '0'
		while len(cur) < k:
			cur = cur + '1' + ''.join(('1' if b == '0' else '0' for b in reversed(cur)))
		return cur[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		memo = {}

		def dp(s):
			if s in memo:
				return memo[s]
			if s == 1:
				return '0'
			temp = dp(s - 1)
			res = temp + '1' + ''.join(['0' if b == '1' else '1' for b in temp])[::-1]
			memo[s] = res
			return res
		return dp(n)[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		curr = '0'
		for x in range(2, n + 1):
			nxt = curr + '1' + self.invert(curr)[::-1]
			curr = nxt
		return curr[k - 1]

	def invert(self, s):
		return ''.join(['1' if x == '0' else '0' for x in s])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def invert(data):
			ans = ''
			for val in data:
				ans += '1' if val == '0' else '0'
			return ans
		output = '0'
		for i in range(1, n + 1):
			output += '1' + invert(output)[::-1]
			if len(output) >= k:
				break
		return output[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s2 = '011'
		if k <= 3:
			return s2[k - 1]
		if k == 2 ** (n - 1):
			return '1'
		elif k < 2 ** (n - 1):
			return self.findKthBit(n - 1, k)
		else:
			ans = self.findKthBit(n - 1, 2 ** n - k)
			return '1' if ans == '0' else '0'


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def helper(res, k):
			if k == 0:
				return res
			invert = ['1' if i == '0' else '0' for i in res]
			invert = ''.join(invert)
			res = res + '1' + invert[::-1]
			return helper(res, k - 1)
		res = helper('0', n - 1)
		return res[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		A = [0]
		while k > len(A):
			A.append(1)
			for i in range(len(A) - 2, -1, -1):
				A.append(A[i] ^ 1)
		return str(A[k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def revinv(x):
			ans = []
			for c in x:
				if c == '0':
					ans.append('1')
				else:
					ans.append('0')
			return ''.join(reversed(ans))
		s = '0'
		while k > len(s):
			s = s + '1' + revinv(s)
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		m = 2 ** n - 1
		if k == 1 or 0 < m - k < 4:
			return '0'
		if k == m or k < 5:
			return '1'
		s = [0]
		for _ in range(n - 1):
			s += [1] + [x ^ 1 for x in reversed(s)]
		return str(s[k - 1])


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = '0'
		if n == 1:
			return '0'
		for i in range(1, n):
			s_i = s[::-1]
			s = s + '1' + ''.join(['1' if c == '0' else '0' for c in s_i])
		return s[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:

		def build_string(n):
			previous = '0'
			for i in range(1, n):
				invert = ''
				for c in previous:
					if c == '1':
						invert += '0'
					else:
						invert += '1'
				previous = previous + '1' + invert[::-1]
			return previous
		ans = build_string(n)
		return ans[k - 1]


class Solution:

	def findKthBit(self, n: int, k: int) -> str:
		s = '011'
		if k < 4:
			return s[k - 1]
		d = {'0': '1', '1': '0'}
		for i in range(n):
			s += '1'
			if len(s) == k:
				return '1'
			else:
				for x in s[:-1][::-1]:
					s += d[x]
					if len(s) == k:
						return d[x]