Java Regex | HackerRank Solution By CodingHumans |

Java Regex

Task 

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

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 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

Hint

1) \\d{1,2} catches any one or two digit number

2) (0|1)\\d{2} catches any three digit number starting with 0 or 1.

3) 2[0-4]\\d catches numbers between 200 and 249.

4) 25[0-5] catches numbers between 250 and 255.

Note that \d represents digits in regular expressions, same as [0-9]

Edit: Replaced "." with "\." as suggested in some of the comments

Recommended: Please try your approach on your integrated development environment (IDE) first, before moving on to the solution.

Few words from CodingHumans : Don't Just copy paste the solution, try to analyze the problem and solve it without looking by taking the the solution as a hint or a reference . Your understanding of the solution matters.

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Solution

( Java )

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{

        public String pattern = "((([01]?[0-9]?[0-9]|2[0-4][0-9]|25[0-5])\\.){3}([01]?[0-9]?[0-9]|2[0-4][0-9]|25[0-5]))\\b";

    }

    

If you have any doubts regarding this problem or  need the solution in other programming languages then leave a comment down below .