# Count of 1-bit and 2-bit characters in the given binary string

Given two special characters, the first character can be represented by one bit which is 0 and the second character can be represented by two bits either 10 or 11. Now given a string represented by several bits. The task is to return the number of characters it represents. Note that the given string is always valid.

Examples:

Input: str = “11100”
Output: 3
“11”, “10” and “0” are the required characters.

Input: str = “100”
Output: 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The approach to solve the problem is that if the current character is 0 then it represents a single character of 1 bit but if the current character is 1 then the next bit after it has to be included in the character consisting of two bits as there is no single bit characters starting with 1.

Below is the implementation of the above approach:

 // C++ implementation of the approach #include 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

Output:
3

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