Count of non-overlapping sub-strings “101” and “010” in the given binary string

Given binary string str, the task is to find the count of non-overlapping sub-strings of either the form “010” or “101”.

Examples:

Input: str = “10101010101” 
Output: 3 
str[0..2] = “101” 
str[3..5] = “010” 
str[6..8] = “101”
Input: str = “111111111111110” 
Output: 0 

Approach: Initialize count = 0 and for every index i in the given string check whether the sub-string of size 3 starting at the current index i matches either with “010” or “101”. If it’s a match then update count = count + 1 and i = i + 3 (to avoid overlapping of sub-strings) else increment i by 1.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the count of
// required non-overlapping sub-strings
int countSubStr(string &s, int n)
{

    // To store the required count
    int count = 0;
    for (int i = 0; i < n - 2;) {

        // If "010" matches the sub-string
        // starting at current index i
        if (s[i] == '0' && s[i + 1] == '1'
            && s[i + 2] == '0') {
            count++;
            i += 3;
        }

        // If "101" matches the sub-string
        // starting at current index i
        else if (s[i] == '1' && s[i + 1] == '0'
                 && s[i + 2] == '1') {
            count++;
            i += 3;
        }
        else {
            i++;
        }
    }

    return count;
}

// Driver code
int main()
{
    string s = "10101010101";
    int n = s.length();

    cout << countSubStr(s, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    // Function to return the count of
    // required non-overlapping sub-strings
    static int countSubStr(char[] s, int n)
    {

        // To store the required count
        int count = 0;
        for (int i = 0; i < n - 2😉
        {

            // If "010" matches the sub-string
            // starting at current index i
            if (s[i] == '0' && s[i + 1] == '1'
                    && s[i + 2] == '0')
            {
                count++;
                i += 3;
            }
            // If "101" matches the sub-string
            // starting at current index i
            else if (s[i] == '1' && s[i + 1] == '0'
                    && s[i + 2] == '1')
            {
                count++;
                i += 3;
            }
            else

            {
                i++;
            }
        }

        return count;
    }

    // Driver code
    public static void main(String[] args)
    {
        char[] s = "10101010101".toCharArray();
        int n = s.length;

        System.out.println(countSubStr(s, n));
    }
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Function to return the count of
# required non-overlapping sub-strings
def countSubStr(s, n) :

    # To store the required count
    count = 0;
    i = 0

    while i < (n-2) :

        # If "010" matches the sub-string
        # starting at current index i
        if (s[i] == '0' and s[i + 1] == '1'and s[i + 2] == '0') :
            count += 1;
            i += 3;

        # If "101" matches the sub-string
        # starting at current index i
        elif (s[i] == '1' and s[i + 1] == '0'and s[i + 2] == '1') :
            count += 1;
            i += 3;

        else :
            i += 1;

    return count;


# Driver code
if __name__ == "__main__" :

    s = "10101010101";
    n = len(s);

    print(countSubStr(s, n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{
    // Function to return the count of
    // required non-overlapping sub-strings
    static int countSubStr(char[] s, int n)
    {

        // To store the required count
        int count = 0;
        for (int i = 0; i < n - 2;)
        {

            // If "010" matches the sub-string
            // starting at current index i
            if (s[i] == '0' &&
                s[i + 1] == '1' &&
                s[i + 2] == '0')
            {
                count++;
                i += 3;
            }

            // If "101" matches the sub-string
            // starting at current index i
            else if (s[i] == '1' &&
                     s[i + 1] == '0' &&
                     s[i + 2] == '1')
            {
                count++;
                i += 3;
            }
            else
            {
                i++;
            }
        }

        return count;
    }

    // Driver code
    public static void Main(String[] args)
    {
        char[] s = "10101010101".ToCharArray();
        int n = s.Length;

        Console.WriteLine(countSubStr(s, n));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// javascript implementation of the approach

    // Function to return the count of
    // required non-overlapping sub-strings
    function countSubStr( s , n) {

        // To store the required count
        var count = 0;
        for (i = 0; i < n - 2;) {

            // If "010" matches the sub-string
            // starting at current index i
            if (s[i] == '0' && s[i + 1] == '1' && s[i + 2] == '0') {
                count++;
                i += 3;
            }
            // If "101" matches the sub-string
            // starting at current index i
            else if (s[i] == '1' && s[i + 1] == '0' && s[i + 2] == '1') {
                count++;
                i += 3;
            } else

            {
                i++;
            }
        }

        return count;
    }

    // Driver code

        var s = "10101010101";
        var n = s.length;

        document.write(countSubStr(s, n));

// This code contributed by Rajput-Ji
</script>

Output:

3

Time Complexity: O(n), where n is the length of the string.
Auxiliary Space: O(1) as constant extra space is used