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  1. Prepare
  2. Java
  3. Strings
  4. Java Substring Comparisons
  5. Discussions

Java Substring Comparisons

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1774 Discussions

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  • amadhuri2006
     + 0 comments

    public class Solution {

    public static void main(String[] args) {
 Scanner sc=new Scanner(System.in);
 String str=sc.next(); 
 int len=sc.nextInt();
  int n=str.length();
 String[] nstr=new String[n-(len-1)];

    for(int i=0;i

    Arrays.sort(nstr); System.out.println(nstr[0]); System.out.println(nstr[n-len]); }

    }

  • ilambrev
     + 0 comments
    import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.nextLine();
        int k = scanner.nextInt();
        scanner.close();
        
        List<String> substrings = new ArrayList<>();
        
        for (int i = 0; i <= s.length() - k; i++) {
            substrings.add(s.substring(i, i + k));
        }
        
        Collections.sort(substrings);
        
        System.out.println(substrings.get(0) + "\n" + substrings.get(substrings.size() - 1));
    }
}

  • cmminsuagrahari
     + 0 comments

    public static String getSmallestAndLargest(String s, int k) { String smallest = ""; String largest = "";

            int n = s.length()-k + 1;
        String[] arr = new String[n];

        for(int i=0; i<=s.length()-k; i++){
                arr[i] = s.substring(i, i+k);
        }

        smallest = arr[0];
        largest = arr[0];
        for(int i=1; i<arr.length; i++){
                if(arr[i].compareTo(smallest) < 0){
                        smallest = arr[i];
                        // largest = arr[i];
                }else if(arr[i].compareTo(largest) > 0){
                        // largest = smallest;
                        largest = arr[i];
                }
        }
        // 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;
}

  • kailash03c
     + 0 comments

    Simple solution

    public static String getSmallestAndLargest(String s, int k) {
        String smallest = s.substring(0 , k);
        String largest = s.substring(0 , k);
        
    
        
        for(int i= 0; i<=s.length()-k; i++){
            String subs  = s.substring(i, i+k);
            
            if(subs.compareTo(smallest)<0){
               smallest = subs; 
            }
              if(subs.compareTo(largest)>0){
               largest = subs; 
            }
            
            
        }
        
        return smallest + "\n" + largest;
    }

  • asrmanalo
     + 0 comments

    import java.util.stream.Collectors; import java.util.stream.IntStream;

    public class Solution { public static String getSmallestAndLargest(String s, int k) {

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

    List<String> subStrings = IntStream.range(0, s.length() - k + 1)
                    .mapToObj(i -> s.substring(i, i + k))
                    .sorted()
                    .collect(Collectors.toList());

    String smallest = subStrings.get(0);
    String largest = subStrings.get(subStrings.size() - 1);

    return smallest + "\n" + largest;
}