Hackerrank Tag Content Extractor Solution

In a tag-based language like XML or HTML, contents are enclosed between a start tag and an end tag like <tag>contents</tag>. Note that the corresponding end tag starts with a /.

Given a string of text in a tag-based language, parse this text and retrieve the contents enclosed within sequences of well-organized tags meeting the following criterion:

1. The name of the start and end tags must be same. The HTML code <h1>Hello World</h2> is not valid, because the text starts with an h1 tag and ends with a non-matching h2 tag.
2. Tags can be nested, but content between nested tags is considered not valid. For example, in <h1><a>contents</a>invalid</h1>, contents is valid but invalid is not valid.
3. Tags can consist of any printable characters.

Input Format

The first line of input contains a single integer,  (the number of lines).
The  subsequent lines each contain a line of text.

Constraints

• Each line contains a maximum of  printable characters.
• The total number of characters in all test cases will not exceed .

Output Format

For each line, print the content enclosed within valid tags.
If a line contains multiple instances of valid content, print out each instance of valid content on a new line; if no valid content is found, print None.

Sample Input

4
<h1>Nayeem loves counseling</h1>
<h1><h1>Sanjay has no watch</h1></h1><par>So wait for a while</par>
<Amee>safat codes like a ninja</amee>
<SA premium>Imtiaz has a secret crush</SA premium>

Sample Output

Nayeem loves counseling
Sanjay has no watch
So wait for a while
None
Imtiaz has a secret crush

Solution in java8

Approach 1.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution{
	public static void main(String[] args){
		
		Scanner in = new Scanner(System.in);
		int testCases = Integer.parseInt(in.nextLine());
		while(testCases>0){
			String line = in.nextLine();
			
          	//Write your code here
			boolean matchFound = false;
            Pattern r = Pattern.compile("<(.+)>([^<]+)</\\1>");
            Matcher m = r.matcher(line);

            while (m.find()) {
                System.out.println(m.group(2));
                matchFound = true;
            }
            if (!matchFound) {
                System.out.println("None");
            }
			testCases--;
		}
	}
}

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Approach 2.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution{
	public static void main(String[] args){
		
		Scanner in = new Scanner(System.in);
		int testCases = Integer.parseInt(in.nextLine());
		while(testCases>0){
			String line = in.nextLine();
			
          	String rgx = "<(.+)>([^<]+)</\\1>";
            Pattern tags = Pattern.compile(rgx);
            Matcher m = tags.matcher(line);
            boolean match = false;
            while (m.find()) {
                System.out.println(m.group(2));
                match = true;
            }
            if (! match == true) {
                System.out.println("None");
            }
			
			testCases--;
		}
	}
}

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Approach 3.

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

public class Solution {
    public static void main(String[] args) {
        Scanner scan  = new Scanner(System.in);
        int testCases = Integer.parseInt(scan.nextLine());
        while (testCases-- > 0) {
            String line = scan.nextLine();
            boolean matchFound = false;
            Pattern r = Pattern.compile("<(.+)>([^<]+)</\\1>");
            Matcher m = r.matcher(line);
            while (m.find()) {
                System.out.println(m.group(2));
                matchFound = true;
            }
            if (!matchFound) {
                System.out.println("None");
            }
        }
        scan.close();
    }
}

		
	

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