Question

write a program to recognize strings under 'ab' '(ab)* '.take input multi time.
do it c/c++/java.

//sample inpute
ab
ababababababab
e(empty)
abaaab
bbaa
//sample output
recognize by 'ab',(ab)*
recognize by '(ab)*'
recognize by '(ab)*'
recognize by '(ab)*'
not recognized
not recognized
help me hurry
​

Answer 1

Explanation:

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

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.

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

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 sumation of 2^ti.

Below is the implementation of above approach.

// 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