import java.io.; import java.util.; import java.util.stream.*;

public class Solution {

public static void main(String[] args) {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */

    try(Scanner s = new Scanner(System.in)){
        //System.out.println("Enter your string");
        String str = s.next();
        //System.out.println("Enter Sub String size or length");
         int len = s.nextInt();

      System.out.println(getSmallestAndLargest(str, len));

    }
}

public static String getSmallestAndLargest(final String str, final int l){
    if(Objects.isNull(str) || str.isBlank() || str.length()< l  || l<=0 || str.length()>1000){
        return "";
    }
   List<String> s = IntStream.range(0, str.length()-l+1)
    .mapToObj(i-> str.substring(i, i+l))
    .map(String::trim)
    .distinct()
    .sorted()
    .collect(Collectors.toList());

    if(s.size() >2){
       return String.format("%s"+"\n"+"%s", s.get(0), s.get(s.size()-1)); 
    }


    if(s.size() ==1){
       return String.format("%s"+"\n"+"%s", s.get(0), s.get(0)); 
    }

    return "";
}

}

** JAVA SOLUTION **

import java.util.Scanner;

public class Solution {

public static String getSmallestAndLargest(String s, int k) {
    String smallest = s.substring(0, k);
    String largest = s.substring(0,k);
    try{
        for (int j = 1; j < s.length()-(k-1); j++) {
            if(s.substring(j, j+k).compareTo(smallest) < 0){
                smallest = s.substring(j, j+k);
            }
            else if(s.substring(j, j+k).compareTo(largest) > 0){
                largest = s.substring(j, j+k);
            }
        }
    }   
    catch (Exception e){
        System.out.println("");
    }

    // Complete the function
    // 'smallest' must be the lexicographically smallest substring of length 'k'
    // 'largest' must be the lexicographically largest substring of length 'k'

    return smallest + "\n" + largest;
}

import java.util.Scanner;

public class Solution {

public static String getSmallestAndLargest(String s, int k) {
    String smallest = s.substring(0, k);
    String largest = s.substring(0, k);

    // Complete the function
    // 'smallest' must be the lexicographically smallest substring of length 'k'
    // 'largest' must be the lexicographically largest substring of length 'k'

    for (int i = 0; i <= s.length() - k; i++) {
      String myWord = s.substring(i, i + k);
      if (smallest.compareTo(myWord) > 0) {
        smallest = myWord;
      }
      if(largest.compareTo(myWord) < 0) {
        largest = myWord;
      }
    }

    return smallest + "\n" + largest;
}


public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    String s = scan.next();
    int k = scan.nextInt();
    scan.close();

    System.out.println(getSmallestAndLargest(s, k));
}

}

Java's Substring Comparisons help find the lexicographically smallest and largest substrings of a given length. It efficiently sorts and compares substrings using string manipulation techniques. Ekbet12

My solution:

public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        String s = sc.nextLine();
        int k = sc.nextInt();
        String smallest = "";
        String largest = "";
        
        sc.close();
        
        ArrayList<String> arrayList = new ArrayList<>();

        for (int i = 0; i < s.length(); i++) {

            if ((i + k) <= s.length()) {
                arrayList.add(s.substring(i, i + k));
            }
        }

        smallest = Collections.min(arrayList);
        largest = Collections.max(arrayList);

        System.out.println(smallest + "\n" + largest);
    }