Java Regex HackerRank Solution

Java Regex HackerRank Solution

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.
```

```000.12.12.034
121.234.12.12
23.45.12.56
```

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

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 StringĀ pattern. 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```

Code Solution:

```#Java Regex HackerRank Solution

import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.Scanner;

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));
}

}
}

class MyRegex{

String zeroTo255
= "(\\d{1,2}|(0|1)\\"
+ "d{2}|2[0-4]\\d|25[0-5])";

String pattern
= zeroTo255 + "\\."
+ zeroTo255 + "\\."
+ zeroTo255 + "\\."
+ zeroTo255;
}

#Java Regex HackerRank Solution```

Disclaimer: This problem is originally created and published by HackerRank, we only provide solutions to this problem. Hence, doesn’t guarantee the truthfulness of the problem. This is only for information purposes.