# Count of operations to make a binary string”ab” free

• Difficulty Level : Hard
• Last Updated : 25 Jul, 2022

Given a string containing characters ‘a’ and ‘b’ only. Convert the given string into a string in which there is no ‘ab’ substring. To make string ‘ab’ free we can perform an operation in which we select a ‘ab’ substring and replace it by ‘bba’.
Find the total number of operations required to convert the given string.

Examples:

```Input : s = 'abbaa'
Output : 2
Explanation:
Here, ['ab'baa] is replaced s = [bbabaa]
[bb'ab'aa] is replaced s = [bbbbaaa]
which is ab free. Hence, 2 operations required.

Input : s = 'aab'
Output : 3
Explanation:
Here, [a'ab'] is replaced s = [abba]
['ab'ba] is replaced s = [bbaba]
[bb'ab'a] is replaced s = [bbbbaa]
which is ab free. Hence, 3 operations required.```

Approach: The final state will be some character ‘a’ after ‘b’: “bbb…baaa…a”
It’s obvious to prove all ‘b’s are distinctive to each other(i.e. Each ‘b’ in the initial state, will add some number of ‘b’s to the final state disjoint from other ‘b’s). For a character ‘b’ from the initial state it will double after seeing a character ‘a’. For each i-th character ‘b’, consider ti the number of a before it. So the final number of ‘b’s can be defined as summation of 2^ti.

Below is the implementation of above approach.

## C++

 `//code to make 'ab' free string` `#include``using` `namespace` `std;` `// code to make 'ab' free string``int` `abFree(string s)``{``    ``int` `n = s.length();``    ``char` `char_array[n + 1];``    ` `    ``// convert string into char array``    ``strcpy``(char_array, s.c_str());``    ` `    ``// Traverse from end. Keep track of count``    ``// b's. For every 'a' encountered, add b_count``    ``// to result and double b_count.``    ``int` `b_count = 0;``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(char_array[n - i - 1] == ``'a'``)``        ``{``            ``res = (res + b_count);``            ``b_count = (b_count * 2);``        ``} ``else` `{``            ``b_count += 1;``        ``}``    ``}``    ``return` `res;``}` `// Driver code``int` `main()``{``    ``string s = ``"abbaa"``;``    ``cout<

## Java

 `//code to make 'ab' free string``import` `java.util.*;` `class` `GFG``{` `    ``// code to make 'ab' free string``    ``static` `int` `abFree(``char``[] s)``    ``{` `        ``// Traverse from end. Keep track of count``        ``// b's. For every 'a' encountered, add b_count``        ``// to result and double b_count.``        ``int` `b_count = ``0``;``        ``int` `res = ``0``;``        ``for` `(``int` `i = ``0``; i < s.length; i++)``        ``{``            ``if` `(s[s.length - i - ``1``] == ``'a'``)``            ``{``                ``res = (res + b_count);``                ``b_count = (b_count * ``2``);``            ``} ``else` `{``                ``b_count += ``1``;``            ``}``        ``}``        ``return` `res;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String s = ``"abbaa"``;``        ``System.out.println(abFree(s.toCharArray()));` `        ``s = ``"aab"``;``        ``System.out.println(abFree(s.toCharArray()));` `        ``s = ``"ababab"``;``        ``System.out.println(abFree(s.toCharArray()));``    ``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# code to make 'ab' free string``def` `abFree(s):``   ` `    ``# Traverse from end. Keep track of count``    ``# b's. For every 'a' encountered, add b_count``    ``# to result and double b_count.``    ``b_count ``=` `0``    ``res ``=` `0``    ``for` `i ``in` `range``(``len``(s)):``        ``if` `s[~i] ``=``=` `'a'``:``            ``res ``=` `(res ``+` `b_count)``            ``b_count  ``=` `(b_count  ``*` `2``)``        ``else``:``            ``b_count  ``+``=` `1``    ``return` `res` `# driver code``s ``=` `'abbaa'``print``(abFree(s))` `s ``=` `'aab'``print``(abFree(s))` `s ``=``'ababab'``print``(abFree(s))`

## C#

 `//code to make 'ab' free string``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``// code to make 'ab' free string``    ``static` `int` `abFree(``char``[] s)``    ``{` `        ``// Traverse from end. Keep track of count``        ``// b's. For every 'a' encountered, add b_count``        ``// to result and double b_count.``        ``int` `b_count = 0;``        ``int` `res = 0;``        ``for` `(``int` `i = 0; i < s.Length; i++)``        ``{``            ``if` `(s[s.Length - i - 1] == ``'a'``)``            ``{``                ``res = (res + b_count);``                ``b_count = (b_count * 2);``            ``} ``else``            ``{``                ``b_count += 1;``            ``}``        ``}``        ``return` `res;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String s = ``"abbaa"``;``        ``Console.WriteLine(abFree(s.ToCharArray()));` `        ``s = ``"aab"``;``        ``Console.WriteLine(abFree(s.ToCharArray()));` `        ``s = ``"ababab"``;``        ``Console.WriteLine(abFree(s.ToCharArray()));``    ``}``}` `// This code contributed by Rajput-Ji`

## Javascript

 ``

Output:

```2
3
11```

Time complexity : O(n)
Auxiliary Space : O(n)

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