每日1題 10/24 找出數組模式(遞規, 0生成01 ,1生成10)

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

 class Solution:

    def kthGrammar(self, n: int, k: int) -> int:

        #由0開始

        if n == 1:

           

            return 0

        if k % 2 == 0:

            #如果K是偶數, 查看上一行的 , 假設k=2, 那就是上前一行k=1處生成,  如果是0 他就是對映1

            return 1 if self.kthGrammar(n-1, k//2) == 0 else 0

        else:

            return 0 if self.kthGrammar(n-1, (k+1)//2) == 0 else 1

       

        # res=[0]*k

        #超出記憶體限制

        # for n in range(n):            

           

        #     array = []

        #     for i in range(k):

        #         # print(f'i={i}, res={res}')

        #         if res[i] == 0:

        #             array.append(0)

        #             array.append(1)

        #         else:

        #             array.append(1)

        #             array.append(0)

        #     res = array

        # # print(f'res={res}')

        # # print(f'array={array}')

        # # print(f'這題的n={n}, k={k}')

        # return array[k-1]

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class Solution:
    def kthGrammar(self, n: int, k: int) -> int:
        if n == 1: return 0
        
        parent = self.kthGrammar(n - 1, ( (k-1) // 2 ) + 1)
        rem = (k - 1) % 2
        
        if parent == 0:
            return 0 if rem == 0 else 1
        else:
            return 1 if rem == 0 else 0