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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

Solution 1

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class Solution:
    def kthGrammar(self, n: int, k: int) -> int:
        if n == 1:
            return 0
        if k <= (1 << (n - 2)):
            return self.kthGrammar(n - 1, k)
        return self.kthGrammar(n - 1, k - (1 << (n - 2))) ^ 1
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class Solution {
    public int kthGrammar(int n, int k) {
        if (n == 1) {
            return 0;
        }
        if (k <= (1 << (n - 2))) {
            return kthGrammar(n - 1, k);
        }
        return kthGrammar(n - 1, k - (1 << (n - 2))) ^ 1;
    }
}
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class Solution {
public:
    int kthGrammar(int n, int k) {
        if (n == 1) return 0;
        if (k <= (1 << (n - 2))) return kthGrammar(n - 1, k);
        return kthGrammar(n - 1, k - (1 << (n - 2))) ^ 1;
    }
};
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func kthGrammar(n int, k int) int {
    if n == 1 {
        return 0
    }
    if k <= (1 << (n - 2)) {
        return kthGrammar(n-1, k)
    }
    return kthGrammar(n-1, k-(1<<(n-2))) ^ 1
}

Solution 2

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class Solution:
    def kthGrammar(self, n: int, k: int) -> int:
        return (k - 1).bit_count() & 1
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class Solution {
    public int kthGrammar(int n, int k) {
        return Integer.bitCount(k - 1) & 1;
    }
}
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class Solution {
public:
    int kthGrammar(int n, int k) {
        return __builtin_popcount(k - 1) & 1;
    }
};
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func kthGrammar(n int, k int) int {
    return bits.OnesCount(uint(k-1)) & 1
}

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