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.
Implementation:
C++
#include <iostream>#include <regex>class GFG{public: static int patternCount(std::string str) { // regular expression for the pattern std::regex regex("10+1"); // compiling regex std::smatch match; // counter int counter = 0; // whenever match found // increment counter while (std::regex_search(str, match, regex)) { // As last character of current match // is always one, starting match from that index str = match.suffix().str(); counter++; } return counter; }};// Driver Method int main(){ std::string str = "1001ab010abc01001"; std::cout << GFG::patternCount(str) << std::endl; return 0;} |
Java
//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)); } } |
Python3
# Python program to count the patterns# of the form 1(0+)1 using Regeximport redef patternCount(str): # regular expression for the pattern regex = "10+1" # compiling regex p = re.compile(regex) # counter counter=0 # whenever match found # increment counter for m in re.finditer(p,str): counter+=1 return counter # Driver Methodstr = "1001ab010abc01001"print(patternCount(str))# This code is contributed by Pushpesh Raj |
C#
// C# program to count the patterns // of the form 1(0+)1 using Regex using System;using System.Text.RegularExpressions;class GFG{ static int patternCount(String str) { // regular expression for the pattern String regex = "10+1"; // compiling regex Regex p = new Regex(regex); // Matcher object Match m = p.Match(str); // counter int counter = 0; // whenever match found // increment counter while(m.Success) { // As last character of current match // is always one, starting match from that index m = m.NextMatch(); counter++; } return counter; } // Driver Method public static void Main (string[] args) { string str = "1001ab010abc01001"; Console.WriteLine(patternCount(str)); } }// This code is contributed by Aman Kumar |
Javascript
// Javascript program for the above approachfunction patternCount(str) { // regular expression for the pattern const regex = /10+1/g; // counter let counter = 0; // whenever match found // increment counter let match; while ((match = regex.exec(str)) !== null) { counter++; } return counter;}// Driver Methodconst str = "1001ab010abc01001";console.log(patternCount(str));// This code is contributed by adityashatmfh |
Output
2
Time Complexity: O(n)
Auxiliary Space: O(1)
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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