Given two strings of equal lengths str1 and str2 consist of characters ‘a’ and ‘b’ only. The following operations can be performed on str1:
- Any character can be changed from ‘a’ to ‘b’ or from ‘b’ to ‘a’ with 1 unit cost.
- Any two characters str1[i] and str1[j] can be swapped with cost |i – j|.
The task is to find the minimum cost required to convert str1 to str2.
Examples:
Input: str1 = “abb”, str2 = “baa”
Output: 2
Explanation: Swap(str1[0], str1[1]), str1 = “bab” and cost = 1
Change str1[2] = ‘b’ to ‘a’, str1 = “baa” and cost = 2Input: str1 = “abab”, str2 = “aabb”
Output: 1
Approach: It can be observed that swapping will only be performed on consecutive characters because if the characters are not consecutive then the cost of swapping will be ? 2 which will give a cost greater than or equal to the cost of changing those characters using the operation of the first type. Now, for every two consecutive characters if they are different in both the string then swap these characters incurring 1 unit cost as compared to 2 unit cost when both of them are changed separately. Else change the character which is different in both the strings (either the current or the next). Finally, print the calculated cost.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the minimum // cost to convert str1 to sr2 int minCost(string str1, string str2, int n) { int cost = 0; // For every character of str1 for ( int i = 0; i < n; i++) { // If current character is not // equal in both the strings if (str1[i] != str2[i]) { // If the next character is also different in both // the strings then these characters can be swapped if (i < n - 1 && str1[i + 1] != str2[i + 1]) { swap(str1[i], str1[i + 1]); cost++; } // Change the current character else { cost++; } } } return cost; } // Driver code int main() { string str1 = "abb" , str2 = "bba" ; int n = str1.length(); cout << minCost(str1, str2, n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the minimum // cost to convert str1 to sr2 static int minCost( char []str1, char []str2, int n) { int cost = 0 ; // For every character of str1 for ( int i = 0 ; i < n; i++) { // If current character is not // equal in both the strings if (str1[i] != str2[i]) { // If the next character is also different in both // the strings then these characters can be swapped if (i < n - 1 && str1[i + 1 ] != str2[i + 1 ]) { swap(str1, i, i + 1 ); cost++; } // Change the current character else { cost++; } } } return cost; } static void swap( char []arr, int i, int j) { char temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } // Driver code public static void main(String[] args) { String str1 = "abb" , str2 = "bba" ; int n = str1.length(); System.out.println(minCost(str1.toCharArray(), str2.toCharArray(), n)); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to return the minimum # cost to convert str1 to sr2 def minCost(str1, str2, n): cost = 0 # For every character of str1 for i in range (n): # If current character is not # equal in both the strings if (str1[i] ! = str2[i]): # If the next character is also different in both # the strings then these characters can be swapped if (i < n - 1 and str1[i + 1 ] ! = str2[i + 1 ]): swap(str1[i], str1[i + 1 ]) cost + = 1 # Change the current character else : cost + = 1 return cost # Driver code if __name__ = = '__main__' : str1 = "abb" str2 = "bba" n = len (str1) print (minCost(str1, str2, n)) # This code is contributed by ashutosh450 |
C#
// C# implementation of the approach using System; class GFG { // Function to return the minimum // cost to convert str1 to sr2 static int minCost( string str1, string str2, int n) { int cost = 0; char [] array = str1.ToCharArray(); // For every character of str1 for ( int i = 0; i < n; i++) { // If current character is not // equal in both the strings if (str1[i] != str2[i]) { // If the next character is also different in both // the strings then these characters can be swapped if (i < n - 1 && str1[i + 1] != str2[i + 1]) { char temp = array[i]; array[i] = array[i + 1]; array[i + 1] = temp ; cost++; } // Change the current character else { cost++; } } } return cost; } // Driver code static public void Main () { string str1 = "abb" , str2 = "bba" ; int n = str1.Length; Console.WriteLine(minCost(str1, str2, n)); } } // This code is contributed by AnkitRai01 |
Javascript
<script> // Javascript implementation of the approach // Function to return the minimum // cost to convert str1 to sr2 function minCost(str1, str2, n) { let cost = 0; let array = str1.split( '' ); // For every character of str1 for (let i = 0; i < n; i++) { // If current character is not // equal in both the strings if (str1[i] != str2[i]) { // If the next character is also different in both // the strings then these characters can be swapped if (i < n - 1 && str1[i + 1] != str2[i + 1]) { let temp = array[i]; array[i] = array[i + 1]; array[i + 1] = temp ; cost++; } // Change the current character else { cost++; } } } return cost; } let str1 = "abb" , str2 = "bba" ; let n = str1.length; document.write(minCost(str1, str2, n)); // This code is contributed by divyeshrabadiya07. </script> |
2
Time Complexity: O(n)
Auxiliary Space: O(1)
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