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This approach utilizes the recursive nature of the transformation of each row. We start from the nth row and trace back to see which position in the previous level gives us the current k-th position.
Time Complexity: O(n), as each recursive call reduces the problem size by a constant factor.
Space Complexity: O(n), due to the recursion stack.
1def kthGrammar(n, k):
2 if n == 1:
3 return 0
4 mid = 1 << (n - 2)
5 if k <= mid:
6 return kthGrammar(n - 1, k)
7 else:
8 return 1 - kthGrammar(n - 1, k - mid)
9
10print(kthGrammar(3, 2)) # Output: 1This function checks whether the index k falls within the first or second half of the row. If it's the first half, it calls itself to check in the (n-1)th row directly. Otherwise, it computes for the corrected position in the (n-1)th row and returns the opposite bit.
The iterative approach to solve this problem is based on tracing the path from the kth element in the nth row back to the root (1st element of the 1st row). Depending on whether k is odd or even, we determine whether to toggle the current result bit as we move up levels until we arrive at the base case.
Time Complexity: O(n).
Space Complexity: O(1), as no additional space is used outside of constant variables.
1using namespace std;
int kthGrammarIterative(int n, int k) {
int result = 0;
while (k > 1) {
if (k % 2 == 0) {
result = 1 - result;
}
k = (k + 1) / 2;
}
return result;
}
int main() {
cout << kthGrammarIterative(3, 2) << endl; // Output: 1
return 0;
}This code snippet iteratively adjusts result based on whether the current index is in the left or the right subtree. At each step, we also adjust k upward until the top of the tree is reached (i.e., k = 1).