Recuento de caracteres de 1 y 2 bits en la string binaria dada

Dados dos caracteres especiales, el primer carácter se puede representar con un bit que es 0 y el segundo carácter se puede representar con dos bits, ya sea 10 u 11 . Ahora dada una string representada por varios bits. La tarea es devolver el número de caracteres que representa. Tenga en cuenta que la string dada siempre es válida.
Ejemplos: 
 

Entrada: str = “11100” 
Salida: 3 
“11”, “10” y “0” son los caracteres requeridos.
Entrada: str = “100” 
Salida: 2 
 

Enfoque: el enfoque para resolver el problema es que si el carácter actual es 0, entonces representa un solo carácter de 1 bit, pero si el carácter actual es 1, entonces el siguiente bit debe incluirse en el carácter que consta de dos bits como no hay caracteres de un solo bit que comiencen con 1.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count
// of required characters
int countChars(string str, int n)
{
 
    int i = 0, cnt = 0;
 
    // While there are characters left
    while (i < n) {
 
        // Single bit character
        if (str[i] == '0')
            i++;
 
        // Two-bit character
        else
            i += 2;
 
        // Update the count
        cnt++;
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    string str = "11010";
    int n = str.length();
 
    cout << countChars(str, n);
 
    return 0;
}

// Java implementation of the above approach
 
class GFG {
     
    // Function to return the count
    // of required characters
    static int countChars(String str, int n)
    {
     
        int i = 0, cnt = 0;
     
        // While there are characters left
        while (i < n) {
     
            // Single bit character
            if (str.charAt(i) == '0')
                i += 1;
     
            // Two-bit character
            else
                i += 2;
     
            // Update the count
            cnt += 1;
        }
     
        return cnt;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        String str = "11010";
        int n = str.length();
     
        System.out.println(countChars(str, n));
    }
    // This code is contributed by AnkitRai01
}

# Python3 implementation of the approach
 
# Function to return the count
# of required characters
def countChars(string, n) :
 
    i = 0; cnt = 0;
 
    # While there are characters left
    while (i < n) :
 
        # Single bit character
        if (string[i] == '0'):
            i += 1;
 
        # Two-bit character
        else :
            i += 2;
 
        # Update the count
        cnt += 1;
 
    return cnt;
 
# Driver code
if __name__ == "__main__" :
 
    string = "11010";
    n = len(string);
 
    print(countChars(string, n));
     
# This code is contributed by AnkitRai01

// C# implementation of the above approach
using System;
 
class GFG
{
     
    // Function to return the count
    // of required characters
    static int countChars(string str, int n)
    {
     
        int i = 0, cnt = 0;
     
        // While there are characters left
        while (i < n)
        {
     
            // Single bit character
            if (str[i] == '0')
                i += 1;
     
            // Two-bit character
            else
                i += 2;
     
            // Update the count
            cnt += 1;
        }
     
        return cnt;
    }
     
    // Driver code
    public static void Main ()
    {
        string str = "11010";
        int n = str.Length;
     
        Console.WriteLine(countChars(str, n));
    }
}
 
// This code is contributed by AnkitRai01

<script>
 
    // JavaScript implementation of the above approach
     
    // Function to return the count
    // of required characters
    function countChars(str, n)
    {
       
        let i = 0, cnt = 0;
       
        // While there are characters left
        while (i < n)
        {
       
            // Single bit character
            if (str[i] == '0')
                i += 1;
       
            // Two-bit character
            else
                i += 2;
       
            // Update the count
            cnt += 1;
        }
       
        return cnt;
    }
     
    let str = "11010";
    let n = str.length;
 
    document.write(countChars(str, n));
     
</script>

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