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
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:
2
Related Articles :
- Regular Expression Java
- Quantifiers
- Extracting each word from a String using Regex
- Check if a given string is a valid number (Integer or Floating Point)
- Print first letter of each word in a string using regex
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