# 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

• 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
}