Find Kth Bit in Nth Binary String

class Solution {
public:
    char findKthBit(int n, int k) {
        int bitPosition = k & -k;
        bool isFlipSection = ((k / bitPosition) >> 1 & 1) == 1;
        bool isFirstBitOne = (k & 1) == 0;

        if (isFlipSection) {
            return isFirstBitOne ? '0' : '1';
        } else {
            return isFirstBitOne ? '1' : '0';
        }
    }
};

The provided C++ code defines a function findKthBit in a class named Solution. This function determines the k-th bit in the n-th iteration of a specific binary string sequence without explicitly generating the strings. The operation mainly revolves around bit manipulation techniques to identify the correct bit.

Here's the logic breakdown:

• bitPosition is calculated using k & -k, which evokes the least significant bit of k that is set to one.
• isFlipSection is deduced using the formula ((k / bitPosition) >> 1 & 1) == 1 that checks if the considered section of the sequence should be flipped based on its position within the sequence.
• isFirstBitOne is determined by (k & 1) == 0 to identify whether the position is originally set to '0' or '1'.

Using these flags, the function decides what the output should be:

• If isFlipSection is true, the result inversely depends on isFirstBitOne.
• Conversely, if isFlipSection is false, the result directly mirrors isFirstBitOne.

This approach effectively sidesteps the need to construct large binary strings, proving efficient especially for high values of n and k.

class Solution {

    public char determineKthBit(int n, int k) {
        // Finding the position of the least significant set bit in k
        int sectionPosition = k & -k;

        // Checking if k is in the part of the string that is inverted
        boolean isInvertedSection = (((k / sectionPosition) >> 1) & 1) == 1;

        // Checking if the bit in its original form before inversion was 1
        boolean bitOriginallyOne = (k & 1) == 0;

        if (isInvertedSection) {
            // Flipping the bit if we are in the inverted section
            return bitOriginallyOne ? '0' : '1';
        } else {
            // Keeping the bit as is if not in the inverted section
            return bitOriginallyOne ? '1' : '0';
        }
    }
}

The provided Java solution tackles the problem of finding the kth bit in the nth binary string (generated recursively). The method determineKthBit is designed to determine the character of the kth bit utilizing bitwise operations without needing to generate the entire string.

Solution Explanation:

• sectionPosition: This finds the position of the least significant set bit in the integer k using the expression k & -k. It identifies which section of the recursively generated string k belongs to.

• isInvertedSection: This boolean determines if the k value is in a part of the binary string that undergoes inversion in the recursive string generation. It uses shift and bitwise operations to ascertain if the section should be inverted ((((k / sectionPosition) >> 1) & 1) == 1).

• bitOriginallyOne: It checks whether the bit in question was originally '1' before any recursive inversion. If the least significant bit of k (k & 1) is 0, then it was originally '1'.

• The final determination of the kth bit considers whether it is in the inverted section:

  • If isInvertedSection is true, it flips the bitOriginallyOne result.
  • If false, it retains the original bit result.

Outcome: The method will return '0' or '1' as the kth bit of the nth binary string. This approach efficiently uses bitwise operations to determine the bit's state directly, which is optimal for large values of n and k, avoiding the potentially vast computational costs of recursively building the string. Note this method depends on the recursive definition and properties of the S sequence the problem refers to.

class Solution:
    def getKthBit(self, seriesLen: int, index: int) -> str:
        # Identify section location for string manipulation based on index
        section_pos = index & -index

        # Check if the current section should be inverted or not
        section_inverted = ((index // section_pos) >> 1 & 1) == 1

        # Verify the original state of the bit
        bit_state = (index & 1) == 0

        # Return the inverted or normal bit state based on section properties
        if section_inverted:
            return "0" if bit_state else "1"
        else:
            return "1" if bit_state else "0"

The provided solution in Python implements a method to find the Kth bit in an algorithmically generated binary string based on a specific series length and the bit's index. The implementation utilizes bitwise operations to efficiently determine and return the state of the Kth bit. Here’s a succinct reiteration of what the code accomplishes:

1. Calculate the position of the section containing the bit for manipulation using bitwise operations. section_pos = index & -index efficiently isolates the lowest set bit of index, which indicates the position.

2. Determine whether the section should be inverted. The expression section_inverted = ((index // section_pos) >> 1 & 1) == 1 checks if the section's order (determined by the ratio of index and section position) is odd or even, defining whether to invert the bit state.

3. Check the original state of the bit without considering section inversion with bit_state = (index & 1) == 0. This step ascertains if the position in a base binary segment is '0' or '1'.

4. Return the correct bit as a string: If the section is inverted, the bit value is flipped; otherwise, it remains the same. This final decision uses the ternary conditions to decide whether the original bit or its inverse should be returned based on the section’s properties.

Essentially, the method effectively leverages mathematical properties and bitwise operations for rapid determination of bit states in a generated binary sequence based on established patterns and series properties.