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
, the1st
row is0
, the2nd
row is01
, and the3rd
row 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 <= 30
1 <= k <= 2n - 1
Companies: Amazon, Bloomberg, Facebook, Adobe, Apple, Google
Related Topics:
Math, Bit Manipulation, Recursion
Hints:
- Try to represent the current (N, K) in terms of some (N-1, prevK). What is prevK ?
// OJ: https://leetcode.com/problems/k-th-symbol-in-grammar/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
int kthGrammar(int n, int k) {
if (n == 1) return 0;
return k <= (1 << (n - 2)) ? kthGrammar(n - 1, k) : (1 - kthGrammar(n - 1, k - (1 << (n - 2))));
}
};
Or
// OJ: https://leetcode.com/problems/k-th-symbol-in-grammar
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
int kthGrammar(int n, int k) {
if (n == 1) return 0;
return kthGrammar(n - 1, (k + 1) / 2) == k % 2;
}
};
// OJ: https://leetcode.com/problems/k-th-symbol-in-grammar/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int kthGrammar(int n, int k) {
int ans = 0;
for (; n > 1; --n) {
if (k > (1 << (n - 2))) {
ans = 1 - ans;
k -= (1 << (n - 2));
}
}
return ans;
}
};
// OJ: https://leetcode.com/problems/k-th-symbol-in-grammar/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
// Ref: https://leetcode.com/problems/k-th-symbol-in-grammar/discuss/113736/PythonJavaC%2B%2B-Easy-1-line-Solution-with-detailed-explanation
class Solution {
public:
int kthGrammar(int N, int K) {
return __builtin_popcount(K - 1) & 1;
}
};