Medium
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
.
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 {
/*
* Time: O(logn)
* Space: O(1)
*/
fun kthGrammar(n: Int, k: Int): Int {
return Integer.bitCount(k - 1) % 2 + n - n
}
}