Find all the patterns of “1(0+)1” in a given string using Regular Expression

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 Regex
import re

def 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 Method

str = "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 approach

function 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 Method
const str = "1001ab010abc01001";
console.log(patternCount(str));

// This code is contributed by adityashatmfh

Output

2

Time Complexity: O(n) 
Auxiliary Space: O(1)

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