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779. K-th Symbol in Grammar

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

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• Java 2
• C++ 2
• Python 2
• Go 2
• TypeScript 2

• class Solution {
      public int kthGrammar(int n, int k) {
          return Integer.bitCount(k - 1) & 1;
      }
  }
  

• class Solution {
  public:
      int kthGrammar(int n, int k) {
          return __builtin_popcount(k - 1) & 1;
      }
  };
  

• class Solution:
      def kthGrammar(self, n: int, k: int) -> int:
          return (k - 1).bit_count() & 1
  
  

• func kthGrammar(n int, k int) int {
  	return bits.OnesCount(uint(k-1)) & 1
  }
  

• function kthGrammar(n: number, k: number): number {
      if (n == 1) {
          return 0;
      }
      if (k <= 1 << (n - 2)) {
          return kthGrammar(n - 1, k);
      }
      return kthGrammar(n - 1, k - (1 << (n - 2))) ^ 1;
  }
  
  

• class Solution {
      public int kthGrammar(int n, int k) {
          return Integer.bitCount(k - 1) & 1;
      }
  }
  

• class Solution {
  public:
      int kthGrammar(int n, int k) {
          return __builtin_popcount(k - 1) & 1;
      }
  };
  

• class Solution:
      def kthGrammar(self, n: int, k: int) -> int:
          return (k - 1).bit_count() & 1
  
  

• func kthGrammar(n int, k int) int {
  	return bits.OnesCount(uint(k-1)) & 1
  }
  

• function kthGrammar(n: number, k: number): number {
      return bitCount(k - 1) & 1;
  }
  
  function bitCount(i: number): number {
      i = i - ((i >>> 1) & 0x55555555);
      i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
      i = (i + (i >>> 4)) & 0x0f0f0f0f;
      i = i + (i >>> 8);
      i = i + (i >>> 16);
      return i & 0x3f;
  }
  
  

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