Find k-th bit in a binary string created by repeated invert and append operations

You are given an initial string s starting with “0”. The string keeps duplicating as follows. Invert of it is appended to it.

Examples:

Input :  k = 2
Output : 1
Initially s = "0".
First Iteration  : s = s + s' = "01"
Second Iteration : s = s + s' = "0110"
The digit at index 2 of s is 1.

Input :  k = 12
Output : 0

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

1. Naive Approach
We can build the string s while its length is smaller than or equal to i in the manner mentioned in the problem description and then simply do a lookup of the required index in the string s.

C++

 // C++ program to find k-th bit in a string // formed by repeated invert and append // operations. #include using namespace std;    int printIndexVal(int k, string s) {     while (s.length() <= k) {            // Building the complement of s         string t = "";         for (int i = 0; i < s.size(); i++) {             if (s[i] == '0')                t += '1';              else                t += '0';         }            // Appending the complement to form         // the new string         s += t;     }        // To match return type     return s[k] - '0'; }    // Driver program to test above function int main() {     string s = "0";     int k = 7;     cout << printIndexVal(k, s) << endl;     return 0; }

Java

 // Java program to find k-th bit in a string  // formed by repeated invert and append  // operations.  class GFG  {    static int printIndexVal(int k, String s)  {      while (s.length() <= k)      {             // Building the complement of s          String t = "";          for (int i = 0; i < s.length(); i++)          {              if (s.charAt(i) == '0')                  t += '1';              else                 t += '0';          }             // Appending the complement to form          // the new string          s += t;      }         // To match return type      return s.charAt(k) - '0';  }     // Driver code  public static void main(String[] args)  {     String s = "0";      int k = 7;      System.out.println(printIndexVal(k, s)); } }     // This code is contributed by 29AjayKumar

C#

 // C# program to find k-th bit in a string  // formed by repeated invert and append  // operations. using System;    class GFG  {    static int printIndexVal(int k, String s)  {      while (s.Length <= k)      {             // Building the complement of s          String t = "";          for (int i = 0; i < s.Length; i++)          {              if (s[i] == '0')                  t += '1';              else                 t += '0';          }             // Appending the complement to form          // the new string          s += t;      }         // To match return type      return s[k] - '0';  }     // Driver code  public static void Main(String[] args)  {     String s = "0";      int k = 7;      Console.WriteLine(printIndexVal(k, s)); } }     // This code contributed by Rajput-Ji

Output:

1

Time Complexity: O(k log k).

Better Approach:
Let’s take a look at a few string constructions:
1. s = 0
2. s = 01
3. s = 0110
4. s = 01101001
5. s = 0110100110010110
Let’s consider finding the bit at position 11 in Line 5. The bit value at this position is effectively the complement of the bit at index 3 in Line 4. So we effectively need to find the complement of the bit at index 3.
s = ~(s). However we do not know s either. Let’s move ahead. Now s = ~(s) using the same explanation in Line 3. And s = ~(s). However, s is always 0 from the problem statement.

Plugging these results,
s = ~(~(~0)) = 1 where ‘~’ is the bitwise NOT operator.
Now, k was initially 11 which is 1011 in binary. The next value of k was 3 that is 011. Notice how the first set bit has reduced from the original k. Subsequently, the next value of k is 1. Another set bit has reduced. Finally for k = 0, the last set bit has vanished. So for k = 11 which is 1011 in binary, the number of complements are 3, which is incidentally equal to the number of set bits in 11. Now we see for odd number of inversions the final result is 1. We can derive the same reasoning for even number of inversions yielding a final result as 0.

To count the number of set-bits in an integer, refer to this article – Count set bits in an integer.

 // C++ program to find k-th bit in a string // formed by repeated invert and append // operations. #include using namespace std;    int printIndexVal(int k) {     // Variable to store set bits     unsigned long long int c = 0;        // Count set bits in a binary number     while (k > 0) {         k &= (k - 1);         c++;     }        // Return 1 if number of set bits     // is odd.     return c & 1; }    // Driver Function int main() {     int k = 12;     cout << printIndexVal(k) << endl; }

Output:

0

Time Complexity: O(log k)

This article is contributed by Aneesh Bose. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
References: https://www.hackerrank.com/contests/w32/challenges/duplication

My Personal Notes arrow_drop_up

Improved By : 29AjayKumar, Rajput-Ji

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.