Regular Expression "[^...]" construct in Java

The subexpression/metacharacter “[^...]” matches any single character, not in brackets.

Example 1

import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class SpecifiedCharacters {
   public static void main( String args[] ) {
      String regex = "[^hwtyoupi]";
      String input = "Hi how are you welcome to Tutorialspoint";
      Pattern p = Pattern.compile(regex);
      Matcher m = p.matcher(input);
      int count = 0;
      while(m.find()) {
         count++;
      }
      System.out.println("Number of matches: "+count);
   }
}

Output

Number of matches: 21

Example 2

The following Java program accepts 5 strings from the user and prints the strings/words that do not contain the English alphabet.

import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexExample {
   public static void main( String args[] ) {
      String regex = "^.*[^a-zA-Z].*$";
      Scanner sc = new Scanner(System.in);
      System.out.println("Enter 5 input strings: ");
      String input[] = new String[5];
      for (int i=0; i
Output
Enter 5 input strings:
1234*5
*&%
sample
test
data23
strings that does not contain English alphabet:
1234*5
*&%