# Time:  O(logn) = O(1) because n is 32-bit integer
# Space: O(1)

# On the first row, we write a 0.
# Now in every subsequent row,
# we look at the previous row and replace each occurrence of 0 with 01,
# and each occurrence of 1 with 10.
#
# Given row N and index K, return the K-th indexed symbol in row N.
# (The values of K are 1-indexed.) (1 indexed).
#
# Examples:
# Input: N = 1, K = 1
# Output: 0
#
# Input: N = 2, K = 1
# Output: 0
#
# Input: N = 2, K = 2
# Output: 1
#
# Input: N = 4, K = 5
# Output: 1
#
# Explanation:
# row 1: 0
# row 2: 01
# row 3: 0110
# row 4: 01101001
#
# Note:
# - N will be an integer in the range [1, 30].
# - K will be an integer in the range [1, 2^(N-1)].

class Solution(object):
    def kthGrammar(self, N, K):
        """
        :type N: int
        :type K: int
        :rtype: int
        """
        def bitCount(n):
            result = 0
            while n:
                n &= n - 1
                result += 1
            return result
        
        return bitCount(K-1) % 2
