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Find all the patterns of “1(0+)1” in a given string (General Approach)

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A string contains patterns of the form 1(0+)1 where (0+) represents any non-empty consecutive sequence of 0’s. Count all such patterns. The patterns are allowed to overlap.

Note : It contains digits and lowercase characters only. The string is not necessarily a binary. 100201 is not a valid pattern. 
One approach to solve the problem is discussed here, other using Regular expressions is given in Set 2 

Examples: 

Input : 1101001
Output : 2

Input : 100001abc101
Output : 2
Recommended Practice

Let size of input string be n. 

  1. Iterate through index ‘0’ to ‘n-1’. 
  2. If we encounter a ‘1’, we iterate till the elements are ‘0’. 
  3. After the stream of zeros ends, we check whether we encounter a ‘1’ or not. 
  4. Keep on doing this till we reach the end of string.

Below is the implementation of the above method. 

Java




// Java Code to count 1(0+)1
// patterns in a string
import java.io.*;
 
class GFG
{
    // Function to count patterns
    static int patternCount(String str)
    {
        /* Variable to store the last character*/
        char last = str.charAt(0);
     
        int i = 1, counter = 0;
        while (i < str.length())
        {
            /* We found 0 and last character was '1',
            state change*/
            if (str.charAt(i) == '0' && last == '1')
            {
                while (i < str.length() && str.charAt(i) == '0')
                    i++;
     
                // After the stream of 0's, we
                // got a '1',counter incremented
                // Break in case of non zero terminating string.
                if (i == str.length()) {
                    break;
                }
                if (str.charAt(i) == '1')
                    counter++;
            }
           
            if (i == str.length()) {
                break;
            }
     
            /* Last character stored */
            last = str.charAt(i);
            i++;
        }
     
        return counter;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        String str = "1001ab010abc01001";
        System.out.println(patternCount(str));
         
    }
}
 
// This code is contributed by vt_m.


C++




/* Code to count 1(0+)1 patterns in a string */
#include <bits/stdc++.h>
using namespace std;
 
/* Function to count patterns */
int patternCount(string str)
{
    /* Variable to store the last character*/
    char last = str[0];
 
    int i = 1, counter = 0;
    while (i < str.size())
    {
        /* We found 0 and last character was '1',
          state change*/
        if (str[i] == '0' && last == '1')
        {
            while (str[i] == '0')
                i++;
 
            /* After the stream of 0's, we got a '1',
               counter incremented*/
            if (str[i] == '1')
                counter++;
        }
 
        /* Last character stored */
        last = str[i];
        i++;
    }
 
    return counter;
}
 
/* Driver Code */
int main()
{
    string str = "1001ab010abc01001";
    cout << patternCount(str) << endl;
    return 0;
}


Python3




# Python3 code to count 1(0+)1 patterns in a
 
# Function to count patterns
def patternCount(str):
     
    # Variable to store the last character
    last = str[0]
 
    i = 1; counter = 0
    while (i < len(str)):
         
        # We found 0 and last character was '1',
        # state change
        if (str[i] == '0' and last == '1'):
            while (str[i] == '0'):
                i += 1
                 
                # After the stream of 0's, we got a '1',
                # counter incremented
                if (str[i] == '1'):
                    counter += 1
         
        # Last character stored
        last = str[i]
        i += 1
     
    return counter
 
 
# Driver Code
str = "1001ab010abc01001"
ans = patternCount(str)
print (ans)
     
# This code is contributed by saloni1297


C#




// C# Code to count 1(0 + )1
// patterns in a string
using System;
 
class GFG
{
     
    // Function to count patterns
    static int patternCount(String str)
    {
        // Variable to store the
        // last character
        char last = str[0];
     
        int i = 1, counter = 0;
        while (i < str.Length)
        {
            // We found 0 and last
            // character was '1',
            // state change
            if (str[i] == '0' && last == '1')
            {
                while (str[i] == '0')
                    i++;
     
                // After the stream of 0's, we
                // got a '1',counter incremented
                if (str[i] == '1')
                    counter++;
            }
     
            // Last character stored
            last = str[i];
            i++;
        }
     
        return counter;
    }
     
    // Driver Code
    public static void Main ()
    {
        String str = "1001ab010abc01001";
        Console.Write(patternCount(str));
         
    }
}
 
// This code is contributed by nitin mittal


PHP




<?php
// PHP Code to count 1(0+)1 patterns
// in a string
 
// Function to count patterns
function patternCount($str)
{
     
    // Variable to store the
    // last character
    $last = $str[0];
 
    $i = 1;
    $counter = 0;
    while ($i < strlen($str))
    {
         
        // We found 0 and last character
        // was '1', state change
        if ($str[$i] == '0' && $last == '1')
        {
            while ($str[$i] == '0')
                $i++;
 
            // After the stream of 0's,
            // we got a '1', counter
            // incremented
            if ($str[$i] == '1')
                $counter++;
        }
 
        /* Last character stored */
        $last = $str[$i];
        $i++;
    }
 
    return $counter;
}
 
    // Driver Code
    $str = "1001ab010abc01001";
    echo patternCount($str) ;
 
// This code is contributed by nitin mittal
?>


Javascript




<script>
 
// javascript Code to count 1(0+)1
// patterns in a string
 
    // Function to count patterns
    function patternCount(str)
    {
        /* Variable to store the last character*/
        var last = str.charAt(0);
     
        var i = 1, counter = 0;
        while (i < str.length)
        {
            /* We found 0 and last character was '1',
            state change*/
            if (str.charAt(i) == '0' && last == '1')
            {
                while (str.charAt(i) == '0')
                    i++;
     
                // After the stream of 0's, we
                // got a '1',counter incremented
                if (str.charAt(i) == '1')
                    counter++;
            }
     
            /* Last character stored */
            last = str.charAt(i);
            i++;
        }
     
        return counter;
    }
     
    // Driver Code
    var str = "1001ab010abc01001";
    document.write(patternCount(str));
 
 
// This code is contributed by 29AjayKumar
</script>


Output

2

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

 



Last Updated : 28 Jan, 2023
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