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<bits/stdc++.h>using namespace std;// code to make 'ab' free stringint 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 codeint 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
//code to make 'ab' free stringimport 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 stringdef 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 codes = 'abbaa'print(abFree(s))s = 'aab'print(abFree(s))s ='ababab'print(abFree(s)) |
C#
//code to make 'ab' free stringusing 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>// code to make 'ab' free stringfunction 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 codevar 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
Time complexity : O(n)
Auxiliary Space : O(n)
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