12 May Java Pattern Syntax Checker – Hacker Rank Solution
Hello Friends, How are you? Today I am going to solve the HackerRank Java Pattern Syntax Checker Problem with a very easy explanation. This is the 20th problem of Java on HackerRank. In this article, you will get more than one approach to solving this problem. So let’s start- {tocify} $title={Table of Contents} Using Regex, we can easily match or search for patterns in a text. Before searching for a pattern, we have to specify one using some well-defined syntax. In this problem, you are given a pattern. You have to check whether the syntax of the given pattern is valid. Note: In this problem, a regex is only valid if you can compile it using the Pattern.compile method. The first line of input contains an integer N, denoting the number of test cases. The next N lines contain a string of any printable characters representing the pattern of a regex. For each test case, print Valid if the syntax of the given pattern is correct. Otherwise, print Invalid. Do not print the quotes.
3 ([A-Z])(.+) [AZ[a-z](a-z) batcatpat(nat {codeBox}
Valid Invalid Invalid {codeBox}
import java.util.Scanner;
import java.util.regex.Pattern;
import java.util.regex.PatternSyntaxException; public class Solution { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int testCases = scan.nextInt(); scan.nextLine(); while (testCases-- > 0) { String pattern = scan.nextLine(); try { Pattern.compile(pattern); System.out.println("Valid"); } catch (PatternSyntaxException exception) { System.out.println("Invalid");
}
}
scan.close();
}
}
Approach II:
import java.util.Scanner;
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 pattern = in.nextLine(); try { Pattern.compile(pattern); System.out.println("Valid"); } catch(PatternSyntaxException e) { System.out.println("Invalid");
}
testCases--;
}
}
}
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