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, the 1st row is 0, the 2nd row is 01, and the 3rd row is 0110.Given two integer n and k, return the kth (1-indexed) symbol in the nth row of a table of n rows.
```

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

- code

```
class Solution:
def kthGrammar(self, n: int, k: int) -> int:
if n == 1:
return 0
# size of each row doubles, so if row 3 has 4 chars, the end idx of that row
# will be 2**(1) == 2
first_half_end_pos = 2**(n - 2)
# if k is in the first half of the str recurse down a row
# if k isn't in the first half of the str, recurse down a row and find the opposite
# char's position in the first half
if k <= first_half_end_pos:
return self.kthGrammar(n - 1, k)
else:
char_in_half_opposite_of_answer = self.kthGrammar(n - 1, k - first_half_end_pos)
return 1 if char_in_half_opposite_of_answer == 0 else
```