Problem
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, the1strow is0, the2ndrow is01, and the3rdrow is0110.
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 <= 301 <= k <= 2n - 1
Solution
/**
* @param {number} n
* @param {number} k
* @return {number}
*/
var kthGrammar = function(n, k) {
var op = 0;
while (n > 1) {
n--;
if (k % 2 === 0) {
op = op === 0 ? 1 : 0;
}
k = Math.ceil(k / 2);
}
return op;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).