How do you solve K-th Symbol in Grammar on LeetCode?

K-th Symbol in Grammar (LeetCode #779) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The sequence grows exponentially, so generating all rows is inefficient.

What is the brute force solution for K-th Symbol in Grammar?

The brute force approach for K-th Symbol in Grammar is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves generating each row of the sequence until we reach the nth row, then simply accessing the k-th symbol. This is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve K-th Symbol in Grammar?

K-th Symbol in Grammar uses the following concepts: Math, Bit Manipulation, Recursion. The recommended patterns to study are: Recursion, Bit Manipulation.

What are the interview tips for K-th Symbol in Grammar?

Understand the problem's recursive structure. Practice breaking down the problem into smaller subproblems. Think about how you can optimize brute force solutions.

What are common mistakes when solving K-th Symbol in Grammar?

Assuming the sequence can be generated efficiently for large n. Not recognizing the recursive nature of the problem.

#779

K-th Symbol in Grammar

Medium

MathBit ManipulationRecursionRecursionBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(1)

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Intuition

Time O(n)Space O(1)

Instead of generating all rows, we can determine the k-th symbol directly using properties of binary representation. The k-th symbol can be derived from the previous row based on the parity of the number of flips.

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Algorithm

3 steps

1Step 1: Calculate the position of the k-th symbol in the previous row using the formula: prevK = (k + 1) / 2.
2Step 2: Check if the number of flips (n - 1) is even or odd to determine the symbol.
3Step 3: If n is odd, return '0' if prevK is even, else return '1'. If n is even, return '1' if prevK is even, else return '0'.

solution.py6 lines

1def kthGrammar(n, k):
2    if n == 1:
3        return 0
4    if k % 2 == 0:
5        return 1 if kthGrammar(n - 1, k // 2) == 0 else 0
6    return kthGrammar(n - 1, (k + 1) // 2)

ℹ

Complexity note: The time complexity is O(n) since we make n recursive calls, but we do not store any intermediate results. The space complexity is O(1) as we only use a constant amount of space.

1The sequence grows exponentially, so generating all rows is inefficient.
2The k-th symbol can be derived recursively without generating the entire sequence.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.