# 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 string` `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` `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 Code` `int` `main()` `{` `    ``int` `N = 4;` `    ``int` `K = 3;`   `    ``cout << findKthChar(N, K);` `    ``return` `0;` `}`

## Java

 `// Java program for ` `// the above approach` `import` `java.util.*;` `class` `GFG{` `  `  `// String reverse` `static` `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 String` `static` `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 Code` `public` `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 string` `def` `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 Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``N ``=` `4` `    ``K ``=` `3`   `    ``print``(findKthChar(N, K))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for ` `// the above approach` `using` `System;` `class` `GFG{` `  `  `// String reverse` `static` `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 String` `static` `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 Code` `public` `static` `void` `Main(String[] args)` `{` `  ``int` `N = 4;` `  ``int` `K = 3;` `  ``Console.Write(findKthChar(N, K));` `}` `}`   `// This code is contributed by Rajput-Ji`

Output:

```B

```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : mohit kumar 29, Rajput-Ji