On the first row, we write a 0. 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.
Given row N and index K, return the K-th indexed symbol in row N. (The values of K are 1-indexed.) (1 indexed).
Examples: Input: N = 1, K = 1 Output: 0 Input: N = 2, K = 1 Output: 0 Input: N = 2, K = 2 Output: 1 Input: N = 4, K = 5 Output: 1 Explanation: row 1: 0 row 2: 01 row 3: 0110 row 4: 01101001
Note:
N will be an integer in the range [1, 30].K will be an integer in the range [1, 2^(N-1)].struct Solution;
impl Solution {
fn kth_grammar(n: i32, k: i32) -> i32 {
let n = n as usize - 1;
let k = k as usize - 1;
Self::kth(n, k)
}
fn kth(n: usize, k: usize) -> i32 {
if n == 0 {
0
} else {
Self::kth(n - 1, k / 2) ^ (k % 2) as i32
}
}
}
#[test]
fn test() {
let n = 1;
let k = 1;
let res = 0;
assert_eq!(Solution::kth_grammar(n, k), res);
let n = 2;
let k = 1;
let res = 0;
assert_eq!(Solution::kth_grammar(n, k), res);
let n = 2;
let k = 2;
let res = 1;
assert_eq!(Solution::kth_grammar(n, k), res);
let n = 4;
let k = 5;
let res = 1;
assert_eq!(Solution::kth_grammar(n, k), res);
let n = 30;
let k = 417_219_134;
let res = 1;
assert_eq!(Solution::kth_grammar(n, k), res);
}