Problem: Lexicographically smallest permutation of a string that can be reduced to length K by removing K-length prefixes from palindromic substrings of length 2K
Given a binary string str of length N, and an integer K, the task is to find the lexicographically smallest permutation of the string str that can be reduced to length K by removal of every K-length prefix from palindromic substrings of length 2K. If no such permutation exists, print "Not Possible".

Sample Input: str = "0000100001100001”, K = 4
Sample Output: 0001100000011000
Explanation: In the string "0001100000011000", every 2K length substring becomes a palindrome whenever a K-length prefix is removed, until string length reduces to K.

Solution (java):
// Java program for the above approach
import java.io.*;

class InfobayAI{

// Function to count the number of
// zeroes present in the string
static int count_zeroes(int n, String str)
{
    int cnt = 0;
    
    // Traverse the string
    for(int i = 0; i < str.length(); i++) 
    {
        if (str.charAt(i) == '0')
            cnt++;
    }

    // Return the count
    return cnt;
}

// Function to rearrange the string s.t
// the string can be reduced to a length
// K as per the given rules
static String kReducingStringUtil(int n, int k, 
                                  String str,
                                  int no_of_zeroes)
{
    
    // Distribute the count of 0s and
    // 1s in segment of length 2k
    int zeroes_in_2k = ((no_of_zeroes) * 
                        (2 * k)) / n;

    int ones_in_2k = 2 * k - zeroes_in_2k;

    // Store string that are initially
    // have formed lexicographically
    // smallest 2k length substring
    String temp_str = "";

    for(int i = 0; i < (zeroes_in_2k) / 2; i++) 
    {
        temp_str += '0';
    }
    for(int i = 0; i < ones_in_2k; i++) 
    {
        temp_str += '1';
    }
    for(int i = 0; i < (zeroes_in_2k) / 2; i++) 
    {
        temp_str += '0';
    }
    
    // Store the lexicographically
    // smallest string of length n
    // that satisfy the condition
    String final_str = "";

    // Insert temp_str into final_str
    // (n/2k) times and add (n%2k)
    // characters of temp_str at end
    for(int i = 0; i < n / (2 * k); i++) 
    {
        final_str += (temp_str);
    }

    for(int i = 0; i < n % (2 * k); i++) 
    {
        final_str += temp_str.charAt(i);
    }

    // Return the final string
    return final_str;
}

// Function to reduce the string to
// length K that follows the given
// conditions
static String kReducingString(int n, int k, 
                              String str)
{
    
    // If the string contains either
    // 0s or 1s then it always be
    // reduced into a K length string
    int no_of_zeroes = count_zeroes(n, str);

    int no_of_ones = n - no_of_zeroes;

    // If the string contains only 0s
    // 1s then it always reduces to
    // a K length string
    if (no_of_zeroes == 0 || 
        no_of_zeroes == n) 
    {
        return str;
    }

    // If K = 1
    if (k == 1) 
    {
        if (no_of_zeroes == 0 || 
            no_of_zeroes == n)
        {
            return str;
        }
        else
        {
            return "Not Possible";
        }
    }

    // Check whether the given string
    // is K reducing string or not
    boolean check = false;

    for(int i = (n / k); i < n; i += (n / k)) 
    {
        if (no_of_zeroes == i || 
            no_of_ones == i)
        {
            check = true;
            break;
        }
    }

    if (check == false)
    {
        return "Not Possible";
    }

    // Otherwise recursively find
    // the string
    return kReducingStringUtil(n, k, str,
                               no_of_zeroes);
}

// Driver Code
public static void main(String[] args)
{
    String str = "0000100001100001";
    int K = 4;
    int N = str.length();

    // Function Call
    System.out.println(kReducingString(N, K, str));
}
}