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

• Difficulty Level : Medium
• Last Updated : 16 Nov, 2022

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:
“11”, “10” and “0” are the required characters.
Input: str = “100”
Output:

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++

 `// 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

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``

Output:

`3`

Time complexity: O(n) where n is length of given binary string
Auxiliary space: O(1) because it is using constant space for variables

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