message: Update k-th-symbol-in-grammar.py
diff: # Time: O(logn)
  # 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):
  """
  