Check if a binary string has a 0 between 1s or not | Set 2 (Regular Expression Approach)

Given a string of 0 and 1, we need to check that the given string is valid or not. The given string is valid when there is no zero is present in between 1's. For example, 1111, 0000111110, 1111000 are valid strings but 01010011, 01010, 101 are not. Examples:

Input : 100
Output : VALID

Input : 1110001
Output : NOT VALID
There is 0 between 1s

Input : 00011
Output : VALID

In Set 1, we have discussed general approach for validity of string.In this post, we will discuss regular expression approach for same and it is simple. As we know that in a string if there is zero between 1's, than string is not valid.Hence below is one of the regular expression for invalid string pattern.

10+1

So here is the simple regex algorithm.

1. Loop over the matcher(string)
2. if above regex match is find in the matcher, then string is not valid, otherwise valid.

Implementation:

C++

// C++ regex program to check for valid string
#include <iostream>
#include <regex>
using namespace std;

    // Method to check for valid string
    bool checkString(string str)
    {
        // regular expression for invalid string
        const regex Regex("10+1");

        // if the regex does not match the string
        // then it is valid
        return !regex_match(str, Regex);
    }

    // Driver method
    int main()
    {
        string str = "00011111111100000";

        // Function call
        cout<<(checkString(str) ? "VALID"
                                        : "NOT VALID");
        
        return 0;
    }

// This code is contributed by Aman Kumar

Java

// Java regex program to check for valid string
  
import java.util.regex.Matcher;
import java.util.regex.Pattern;
  
class GFG 
{
    // Method to check for valid string
    static boolean checkString(String str)
    {
        // regular expression for invalid string
        String regex = "10+1";
          
        // compiling regex
        Pattern p = Pattern.compile(regex);
          
        // Matcher object
        Matcher m = p.matcher(str);
          
        // loop over matcher
        while(m.find())
        {
            // if match found,
            // then string is invalid
            return false;
        }
           
        // if match doesn't found
        // then string is valid
        return true;    
    }
  
    // Driver method
    public static void main (String[] args)
    {
        String str = "00011111111100000";
          
        System.out.println(checkString(str) ? "VALID" : "NOT VALID");
    }
}

Python3

# Python3 regex program to check for valid string
import re

# Method to check for valid string
def checkString(str):
    # regular expression for invalid string
    regex = "10+1"
    x = re.search("10+1", str)
    return x is None

#Driver method
str = "00011111111100000"

if checkString(str):
    print("VALID")
else:
    print("NOT VALID")
    
#this code is contributed by phasing17

C#

// C# regex program to check for valid string

using System;
using System.Text.RegularExpressions;

class GFG {
    // Method to check for valid string
    static bool checkString(string str)
    {
        // regular expression for invalid string
        Regex regex = new Regex(@"10+1");

        // if the regex does not match the string
        // then it is valid
        return !regex.IsMatch(str);
    }

    // Driver method
    public static void Main(string[] args)
    {
        string str = "00011111111100000";

        // Function call
        Console.WriteLine(checkString(str) ? "VALID"
                                           : "NOT VALID");
    }
}

// This code is contributed by phasing17

JavaScript

// JavaScript regex program to check for valid string


// Method to check for valid string
function checkString(str)
{
    // returns true if string contains only 0 and 1
    // otherwise, returns false
    return str.match(/^[0-1]+$/) != null;

}

  
// Driver method
let str = "01";
          
console.log(checkString(str) ? "VALID" : "NOT VALID");


// This code is contributed by phasing17

Output

VALID

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

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