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

  • Difficulty Level : Hard
  • Last Updated : 28 May, 2021

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.

 

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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++




// C++ program to find all subsets of given set. Any
// repeated subset is considered only once in the output
#include<bits/stdc++.h>
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<<abFree(s)<<endl;
     
    s = "aab";
    cout<<abFree(s)<<endl;
     
    s = "ababab";
    cout<<abFree(s)<<endl;
     
    return 0;
}
 
// This code is contributed by Rajput-Ji

Java




// Java program to find all subsets of given set. Any
// repeated subset is considered only once in the output
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#




// C# program to find all subsets of given set.
// Any repeated subset is considered
// only once in the output
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




<script>
 
// JavaScript program to find
// all subsets of given set. Any
// repeated subset is considered
// only once in the output
 
// code to make 'ab' free string
function abFree(s)
{
    var n = s.length;
     
    // convert string into char array
    var char_array = s.split('')
     
    // Traverse from end. Keep track of count
    // b's. For every 'a' encountered, add b_count
    // to result and double b_count.
    var b_count = 0;
    var res = 0;
    for (var 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
var s = "abbaa";
document.write( abFree(s) + "<br>");
 
s = "aab";
document.write( abFree(s) + "<br>");
 
s = "ababab";
document.write( abFree(s) + "<br>");
 
 
</script>

Output:  

2
3
11

 




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