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# Kth character from the Nth string obtained by the given operations

Given two positive integers N and K, the task is to find the Kth character of a string obtained by performing the following operation on a string S( initially “A”) N times.

Every ith operation generates following string (Si):
Si = Si – 1 + ‘B’ + rev(comp(Si – 1))
where,
comp() denotes the complement of string i.e., A is changed to B and B is changed to A
and rev() returns the reverse of a string.

Examples:

Input: N = 3, K = 1
Output:
Explanation:
Initially, after first operation, S1 = “A”
After 2nd operation, S2 = “ABB”
After 3rd operation, S3 = “ABBBAAB”
The first character of S3 is ‘A’.
Input: N = 2, K = 3
Output: B

Approach: The idea is to use recursion to generate the new string from the previously generated string and to repeat the process until N operations are performed. Below are the steps:

1. Initialize two string prev as “A” and curr as an empty string.
2. If N = 1 then returns prev otherwise perform below operation.
3. Iterate a loop (N – 1) times, each time update curr as prev + “B”.
4. Reverse the string prev.
5. Again update the string curr string as curr + prev.
6. Update string prev as curr.
7. After the above steps return the (K – 1)th character of the string curr.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std; // Function to return Kth character// from recursive stringchar findKthChar(int n, int k){    string prev = "A";    string cur = "";     // If N is 1 then return A    if (n == 1) {        return 'A';    }     // Iterate a loop and generate    // the recursive string    for (int i = 2; i <= n; i++) {         // Update current string        cur = prev + "B";         // Change A to B and B to A        for (int i = 0;             i < prev.length(); i++) {             if (prev[i] == 'A') {                prev[i] = 'B';            }            else {                prev[i] = 'A';            }        }         // Reverse the previous string        reverse(prev.begin(), prev.end());        cur += prev;        prev = cur;    }     // Return the kth character    return cur[k - 1];} // Driver Codeint main(){    int N = 4;    int K = 3;     cout << findKthChar(N, K);    return 0;}

## Java

 // Java program for// the above approachimport java.util.*;class GFG{   // String reversestatic String reverse(String input){  char[] a = input.toCharArray();  int l, r = a.length - 1;  for (l = 0; l < r; l++, r--)  {    char temp = a[l];    a[l] = a[r];    a[r] = temp;  }  return String.valueOf(a);}   // Function to return Kth character// from recursive Stringstatic char findKthChar(int n,                        int k){  String prev = "A";  String cur = "";   // If N is 1 then return A  if (n == 1)  {    return 'A';  }   // Iterate a loop and generate  // the recursive String  for (int j = 2; j <= n; j++)  {    // Update current String    cur = prev + "B";     // Change A to B and B to A    for (int i = 0; i < prev.length(); i++)    {      if (prev.charAt(i) == 'A')      {        prev.replace(prev.charAt(i), 'B');      }      else      {        prev.replace(prev.charAt(i), 'A');      }    }     // Reverse the previous String    prev = reverse(prev);    cur += prev;    prev = cur;  }   // Return the kth character  return cur.charAt(k);} // Driver Codepublic static void main(String[] args){  int N = 4;  int K = 3;  System.out.print(findKthChar(N, K));}} // This code is contributed by Rajput-Ji

## Python3

 # Python3 program for the above approach # Function to return Kth character# from recursive stringdef findKthChar(n, k):     prev = "A"    cur = ""     # If N is 1 then return A    if (n == 1):        return 'A'     # Iterate a loop and generate    # the recursive string    for i in range(2, n + 1):         # Update current string        cur = prev + "B"         # Change A to B and B to A        temp1 = [y for y in prev]                 for i in range(len(prev)):            if (temp1[i] == 'A'):                temp1[i] = 'B'            else:                temp1[i] = 'A'         # Reverse the previous string        temp1 = temp1[::-1]        prev = "".join(temp1)        cur += prev        prev = cur     # Return the kth character    return cur[k - 1] # Driver Codeif __name__ == '__main__':         N = 4    K = 3     print(findKthChar(N, K)) # This code is contributed by mohit kumar 29

## C#

 // C# program for// the above approachusing System;class GFG{   // String reversestatic String reverse(String input){  char[] a = input.ToCharArray();  int l, r = a.Length - 1;  for (l = 0; l < r; l++, r--)  {    char temp = a[l];    a[l] = a[r];    a[r] = temp;  }  return String.Join("", a);}   // Function to return Kth character// from recursive Stringstatic char findKthChar(int n,                        int k){  String prev = "A";  String cur = "";   // If N is 1 then return A  if (n == 1)  {    return 'A';  }   // Iterate a loop and generate  // the recursive String  for (int j = 2; j <= n; j++)  {    // Update current String    cur = prev + "B";     // Change A to B and B to A    for (int i = 0; i < prev.Length; i++)    {      if (prev[i] == 'A')      {        prev.Replace(prev[i], 'B');      }      else      {        prev.Replace(prev[i], 'A');      }    }     // Reverse the previous String    prev = reverse(prev);    cur += prev;    prev = cur;  }   // Return the kth character  return cur[k];} // Driver Codepublic static void Main(String[] args){  int N = 4;  int K = 3;  Console.Write(findKthChar(N, K));}} // This code is contributed by Rajput-Ji

## Javascript



Output:

B

Time Complexity: O(N2)
Auxiliary Space: O(1)