How do you solve Find Kth Bit in Nth Binary String on LeetCode?

Find Kth Bit in Nth Binary String (LeetCode #1545) 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 middle bit of S_n is always '1'.

What is the brute force solution for Find Kth Bit in Nth Binary String?

The brute force approach for Find Kth Bit in Nth Binary String is Brute Force, with O(n²) time complexity and O(n) space complexity. We can construct the binary string S_n directly by following the recursive definition. Although this approach is straightforward, it can be inefficient for larger values of n due to the exponential growth of the string length. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Kth Bit in Nth Binary String?

Find Kth Bit in Nth Binary String uses the following concepts: String, Recursion, Simulation. The recommended patterns to study are: Recursion, Divide and Conquer.

What are the interview tips for Find Kth Bit in Nth Binary String?

Always consider if you can avoid constructing large data structures. Think about the properties of the problem that can simplify your approach. Practice breaking down recursive problems into smaller subproblems.

What are common mistakes when solving Find Kth Bit in Nth Binary String?

Confusing the indices when accessing characters in the string. Not recognizing the recursive nature of the problem and trying to build the entire string.

#1545

Find Kth Bit in Nth Binary String

Medium

StringRecursionSimulationRecursionDivide and Conquer

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of constructing the entire string, we can use properties of the binary string's structure to find the k-th bit directly. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Determine the length of S_n, which is 2^n - 1.
2Step 2: If k is equal to the middle index (2^(n-1)), return '1'.
3Step 3: If k is less than the middle index, recursively find the k-th bit in S_(n-1).
4Step 4: If k is greater than the middle index, find the corresponding bit in the inverted part and return its inverted value.

solution.py10 lines

1def findKthBit(n, k):
2    if n == 1:
3        return '0'
4    mid = 2**(n - 1)
5    if k == mid:
6        return '1'
7    elif k < mid:
8        return findKthBit(n - 1, k)
9    else:
10        return '1' if findKthBit(n - 1, 2 * mid - k) == '0' else '0'

ℹ

Complexity note: The time complexity is O(n) because we only make a constant number of recursive calls proportional to n, without constructing large strings. The space complexity is O(1) since we only use a few variables for calculations.

1The middle bit of S_n is always '1'.
2The structure of S_n allows us to avoid constructing the entire string.

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