Description
We build a table of n rows (1-indexed). We start by writing 0 in the 1st row. 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.
- For example, for 
n = 3, the1strow is0, the2ndrow is01, and the3rdrow is0110. 
Given two integer n and k, return the kth (1-indexed) symbol in the nth row of a table of n rows.
Β
Example 1:
Input: n = 1, k = 1 Output: 0 Explanation: row 1: 0
Example 2:
Input: n = 2, k = 1 Output: 0 Explanation: row 1: 0 row 2: 01
Example 3:
Input: n = 2, k = 2 Output: 1 Explanation: row 1: 0 row 2: 01
Β
Constraints:
1 <= n <= 301 <= k <= 2n - 1
Solution
Python3
class Solution:
    def kthGrammar(self, n: int, k: int) -> int:
        
        @cache
        def go(n, k):
            if n == 1: return 0
 
            # right child
            if k % 2 == 0:
                if go(n - 1, k // 2) == 0: 
                    return 1
                else:
                    return 0
            else:
                if go(n - 1, (k + 1) // 2) == 0:
                    return 0
                else:
                    return 1
        
        return go(n, k)