Find all the patterns of "1(0+)1" in a given string | SET 2(Regular Expression Approach)

Find all the patterns of “1(0+)1” in a given string | SET 2(Regular Expression Approach)

Difficulty Level : Easy
Last Updated : 30 May, 2018

In Set 1, we have discussed general approach for counting the patterns of the form 1(0+)1 where (0+) represents any non-empty consecutive sequence of 0’s.In this post, we will discuss regular expression approach to count the same.

Examples:

Input : 1101001
Output : 2

Input : 100001abc101
Output : 2

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Below is one of the regular expression for above pattern

10+1

Hence, whenever we found a match, we increase counter for counting the pattern.As last character of a match will always ‘1’, we have to again start searching from that index.

//Java program to count the patterns  
// of the form 1(0+)1 using Regex 
  
import java.util.regex.Matcher; 
import java.util.regex.Pattern; 
  
class GFG  
{ 
    static int patternCount(String str)  
    { 
        // regular expression for the pattern 
        String regex = "10+1"; 
          
        // compiling regex 
        Pattern p = Pattern.compile(regex); 
                  
        // Matcher object 
        Matcher m = p.matcher(str); 
          
        // counter 
        int counter = 0; 
                  
        // whenever match found 
        // increment counter 
        while(m.find()) 
        { 
            // As last character of current match 
            // is always one, starting match from that index 
            m.region(m.end()-1, str.length()); 
              
            counter++; 
        } 
                  
        return counter; 
    } 
      
    // Driver Method 
    public static void main (String[] args) 
    { 
        String str = "1001ab010abc01001"; 
        System.out.println(patternCount(str)); 
    } 
} 

Output:

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