Problem Description

The problem requires us to generate a binary string Sn using a specified algorithm and then find the kth bit in this string for the given n. The binary string is constructed following these rules:

• Start with S1 as "0".
• For any i > 1, the string Si is formed by concatenating the previous string Si-1 with "1" and then adding the reversed and inverted version of Si-1.

To clarify, the invert function flips all bits in a string (0 becomes 1 and 1 becomes 0), and the reverse function reverses the order of the characters in the string.

A simple illustration shows that the strings grow exponentially as n increases and the process of creating them is recursive.

The challenge is to find the kth bit in the constructed string without actually building the entire string, as doing so would be highly inefficient for large n.

Intuition

The provided solution navigates through the structure of the generated strings using the defined construction algorithm without having to construct the entire string Sn. Here's how it achieves this:

1. The length of the final string Sn is determined by a recursive function calcLength. Since the length is doubled and incremented by 1 with each iteration, this function keeps track and uses a set to record specific positions that would contain the bit "1".

2. Base cases are checked: if either n is 1 or k is 1, it's clear from the construction rules that the kth bit must be "0".

3. Knowing that the string is symmetric with a "1" in the middle for any n > 1, there are three cases:

   • If k is at the midpoint, return "1".
   • If k is before the midpoint, recursively find the kth bit in Sn-1.
   • If k is after the midpoint, find the corresponding bit before the midpoint in Sn-1 and invert it.

4. A helper function r is used to invert a given bit.

The recursive strategy works by leveraging symmetry and the known structure of the string, avoiding unnecessary computation of the full string and thus maintaining efficiency even for large n.

Learn more about Recursion patterns.

Solution Approach

The solution utilizes a recursive approach, coupled with a set for bookkeeping, to simplify the task of finding the kth bit in the sequence Sn. The implementation is structured as follows:

• The key to the solution lies in understanding the symmetry of the string Sn. Each string Si for i > 1 is a palindrome with a center bit '1'. This property allows us to only focus on half of the string to find the corresponding value.

• The calcLength function calculates the length of Sn recursively based on the formula that the length of Si is twice the length of Si-1 plus one (len(Si) = 2 * len(Si-1) + 1). The function also adds the length len(Si) plus one to the set set for each iteration, marking positions in the string which result in the bit '1'.

• The findKthBit function acts as the primary function to determine the kth bit:

  • Base cases check if k == 1 or n == 1, and since we know the first character is always '0', it returns '0'.
  • If the kth position has been marked in the set (which was done in calcLength), it means that the kth bit is '1'.
  • Otherwise, the function determines if the kth bit is before or after the midpoint of Sn:

    • If k is less than the midpoint, recursion is performed to find the kth bit in Sn-1.

    • If k is at or beyond the midpoint:

      • The midpoint bit is always '1', which is directly returned if k is exactly the midpoint.
      • For bits after the midpoint, it must find the corresponding bit before the midpoint in Sn-1 and invert it using the r function.

• The r function is the inverting function, which takes a bit character and returns its inversion.

This implementation forgoes the need to construct the entire Sn string by smartly exploiting its properties, specifically the palindrome characteristic and marking of certain positions with known bits. The algorithm's efficiency arises from avoiding the explicit generation of the binary string and instead working through recursive calls that "simulate" the construction process minimally to find the required bit.

Example Walkthrough

Let's illustrate the solution approach using a small example by constructing the string S3 and finding the 5th bit of S3.

Starting with S1 as "0" and following the problem rules, we can construct the strings S2 and S3:

• S1 is "0"
• To get S2, we take S1, concat a "1", and then add the inverted and reversed S1: "0 1 1" (where the space indicates the midpoint)
• For S3, the process is repeated with S2: "011 1 100" (again, space shows the midpoint)

The length of S3 is 7 bits, and it is a palindrome with the 4th bit being the middle "1". We can visualize it as: [0] [1] [1] [1] [0] [0] [0] where brackets show the positions: 1 2 3 4 5 6 7

We want to find the 5th bit:

1. As per the rule, the midpoint of S3 is bit 4 and is "1". So we split the string into two halves around the midpoint:
   • First half (left of midpoint): [0] [1] [1]
   • Second half (right of midpoint): [0] [0] [0]
2. We see that the 5th bit is on the right side of the midpoint, which is the reverse and inverted of the first half. So the bit corresponding to the 5th bit is the 3rd bit from the first half (S2).
3. Now we invert that bit. The 3rd bit of S2 is "1" (since S2 is "011"), so its inversion is "0".
4. We conclude that the 5th bit of S3 is "0".

This example highlights the application of known symmetry and inversion to determine the value of the kth bit without constructing the full string Sn, demonstrating the algorithm's efficiency.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the findKthBit function involves multiple recursive calls. The calcLength function is called recursively to calculate the length of the Sn string. Due to the nature of the construction of Sn, where Sn = Sn-1 + "1" + reverse(invert(Sn-1)), the length of each subsequent string is double the previous plus one, making it an exponential growth in terms of n.

Let's denote the time complexity of calcLength as T(n). Therefore, we have:

T(n) = T(n-1) + O(1)

Since this recurses n times, and each recursion is just adding a constant amount of work (besides the recursive call), this simplifies to a linear complexity in terms of n:

T(n) = O(n)

Next, analyzing the recursive calls in the main findKthBit function, we notice that in the worst case, it can end up recursively calling itself by decreasing n by one each time until n reaches 1. This gives us, in the worst case, a recursion depth of n.

However, we also need to consider that not every recursive call goes all the way down to n=1 due to the early return conditions, so not every call will branch out fully. The recursion only happens to find the kth bit before the midpoint. After the midpoint, the work reduces due to the reuse of the computation as we call findKthBit(n - 1, len - k + 1).

As a result, the time complexity is not straightforward to calculate, but we can approximate it by considering that at each level of recursion, the size of the problem is roughly halved (similar to a binary search). The worst-case time complexity thus approximates to:

O(n * log(len))

where len is the length of the final string Sn which is O(2^n).

Combining all the recursive calls and operations within each call, the total time complexity roughly approximates to:

O(n^2)

This is considering that each recursive call to findKthBit has a cost that doubles each time with a diminishing number of calls due to halving of k as well.

Space Complexity

The space complexity is governed by both the recursive call stack and the space taken by the HashSet to store specific lengths at each level of recursion.

The maximum depth of the recursion stack is n, for the recursive calls to findKthBit.

The HashSet grows in size linearly with n, since a new length is added for each level of recursion in calcLength.

Therefore, the space complexity is:

O(n) + O(n) = O(n)

Overall, the space complexity of the function is linear with respect to n due to the recursion stack and the HashSet storage:

O(n)

Learn more about how to find time and space complexity quickly using problem constraints.