Problem Link

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, 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

Example 3:

Input: n = 2, k = 2
Output: 1
Explanation: 
row 1: 0
row 2: 01

 

Constraints:

  • 1 <= n <= 30
  • 1 <= 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)