Java Regex Hackerrank Problem Solution

Java Regex Hackerrank

Java Regex Hackerrank

Explanation:-

Write a class called MyRegex which will contain a string pattern. You need to write a regular expression and assign it to the pattern such that it can be used to validate an IP address. Use the following definition of an IP address:

IP address is a string in the form “A.B.C.D”, where the value of A, B, C, and D may range from 0 to 255. Leading zeros are allowed. The length of A, B, C, or D can’t be greater than 3.

Some valid IP address:

000.12.12.034
121.234.12.12
23.45.12.56
Some invalid IP address:

000.12.234.23.23
666.666.23.23
.213.123.23.32
23.45.22.32.
I.Am.not.an.ip

 

:- Write a class called MyRegex which will contain a string pattern. You need to write a regular expression and assign it to the pattern such that it can be used to validate an IP address.

In this problem, you will be provided strings containing any combination of ASCII characters. You have to write a regular expression to find the valid IPs.

Just write the MyRegex class which contains a String. The string should contain the correct regular expression.

(MyRegex class MUST NOT be public)

Sample Input

000.12.12.034
121.234.12.12
23.45.12.56
00.12.123.123123.123
122.23
Hello.IP
Sample Output

True 
true 
true 
false 
false 
false
Solution

 

import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.*;
class Solution{

public static void main(String[] args){
Scanner in = new Scanner(System.in);
while(in.hasNext()){
String IP = in.next();
System.out.println(IP.matches(new MyRegex().pattern));
}

}
}
//Write your code here
  class MyRegex {
      String pattern;
    MyRegex() {     
        String zeroTo255 ="(\\d{1,2}|(0|1)\\d{2}|2[0-4]\\d|25[0-5])";
        pattern = zeroTo255 + "." + zeroTo255 + "." + zeroTo255 + "." + zeroTo255;

    }
}

 

Just write the MyRegex class which contains a String. The string should contain the correct regular expression.

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