Given two strings A and B of length N and M respectively, the task is to find the minimum cost to convert string A to B using the following operations:
- A character of string A can be swapped from another character of the same string. Cost = 0.
- A character can be deleted from string B or can be inserted in the string A. Cost = 1.
Examples:
Input: A = “1aB+-“, B = “CC”
Output: 7
Explanation: Remove all 5 characters from string A and insert character C twice. Therefore, the total cost = 5 + 2 = 7.Input: A = “aBcD”, B = “DBac”
Output: 0
Explanation: Following operations need to be performed to convert string A to string B:
- Swap ‘a’ with ‘D’. Therefore, A = “DBca”.
- Swap ‘a’ with ‘c’. Therefore, A = “DBac”.
Therefore, the total cost = 0.
Approach: The idea is to perform a swap operation maximum number of times to reduce the total cost. Observe that the characters which are common between the strings A and B can be swapped any number of times in A to make the string equal to B. All the characters that are present in the string A but not in the string B have to be deleted from A and all the characters present in B and not present in A have to be inserted in A to make both the strings equal. Follow the steps below to solve the problem:
- Initialize two arrays a[] and b[] of length 256 to store the frequencies of each character in the strings A and B respectively.
- Initialize a variable, say minCost, to store the minimum cost.
- Traverse over the range [0, 255] using the variable i and at each iteration, increment minCost by abs(a[i] – b[i]).
- After completing the above steps, print the value of minCost as the minimum cost required to convert string A to B.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the minimum cost // to convert string a to b void minimumCost(string a, string b)
{ // Stores the frequency of string
// a and b respectively
vector< int > fre1(256), fre2(256);
// Store the frequencies of
// characters in a
for ( char c : a)
fre1[( int )(c)]++;
// Store the frequencies of
// characters in b
for ( char c : b)
fre2[( int )(c)]++;
// Minimum cost to convert A to B
int mincost = 0;
// Find the minimum cost
for ( int i = 0; i < 256; i++) {
mincost += abs (fre1[i]
- fre2[i]);
}
// Print the minimum cost
cout << mincost << endl;
} // Driver Code int main()
{ string A = "1AB+-" , B = "cc" ;
// Function Call
minimumCost(A, B);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the minimum cost // to convert string a to b public static void minimumCost(String a, String b)
{ // Stores the frequency of string
// a and b respectively
int fre1[] = new int [ 256 ];
int fre2[] = new int [ 256 ];
// Store the frequencies of
// characters in a
for ( char c : a.toCharArray())
fre1[( int )(c)]++;
// Store the frequencies of
// characters in b
for ( char c : b.toCharArray())
fre2[( int )(c)]++;
// Minimum cost to convert A to B
int mincost = 0 ;
// Find the minimum cost
for ( int i = 0 ; i < 256 ; i++)
{
mincost += Math.abs(fre1[i] -
fre2[i]);
}
// Print the minimum cost
System.out.println(mincost);
} // Driver Code public static void main(String[] args)
{ String A = "1AB+-" , B = "cc" ;
// Function Call
minimumCost(A, B);
} } // This code is contributed by divyeshrabadiya07 |
# Python3 program for the above approach # Function to find the minimum cost # to convert a to b def minimumCost(a, b):
# Stores the frequency of string
# a and b respectively
fre1 = [ 0 ] * ( 256 )
fre2 = [ 0 ] * ( 256 )
# Store the frequencies of
# characters in a
for c in a:
fre1[ ord (c)] + = 1
# Store the frequencies of
# characters in b
for c in b:
fre2[ ord (c)] + = 1
# Minimum cost to convert A to B
mincost = 0
# Find the minimum cost
for i in range ( 256 ):
mincost + = abs (fre1[i] - fre2[i])
# Print the minimum cost
print (mincost)
# Driver Code if __name__ = = '__main__' :
A = "1AB+-"
B = "cc"
# Function Call
minimumCost(A, B)
# This code is contributed by mohit kumar 29 |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimum cost // to convert string a to b public static void minimumCost( string a,
string b)
{ // Stores the frequency of string
// a and b respectively
int [] fre1 = new int [256];
int [] fre2 = new int [256];
// Store the frequencies of
// characters in a
foreach ( char c in a.ToCharArray())
fre1[( int )(c)]++;
// Store the frequencies of
// characters in b
foreach ( char c in b.ToCharArray())
fre2[( int )(c)]++;
// Minimum cost to convert A to B
int mincost = 0;
// Find the minimum cost
for ( int i = 0; i < 256; i++)
{
mincost += Math.Abs(fre1[i] -
fre2[i]);
}
// Print the minimum cost
Console.Write(mincost);
} // Driver code public static void Main()
{ string A = "1AB+-" , B = "cc" ;
// Function Call
minimumCost(A, B);
} } // This code is contributed by sanjoy_62 |
<script> // Javascript program for the above approach // Function to find the minimum cost // to convert string a to b function minimumCost(a, b)
{ // Stores the frequency of string
// a and b respectively
var fre1 = Array(256).fill(0), fre2= Array(256).fill(0);
// Store the frequencies of
// characters in a
a.split( '' ).forEach(c => {
fre1[c.charCodeAt(0)]++;
});
// Store the frequencies of
// characters in b
b.split( '' ).forEach(c => {
fre2[c.charCodeAt(0)]++;
});
// Minimum cost to convert A to B
var mincost = 0;
// Find the minimum cost
for ( var i = 0; i < 256; i++) {
mincost += Math.abs(fre1[i]
- fre2[i]);
}
// Print the minimum cost
document.write( mincost );
} // Driver Code var A = "1AB+-" , B = "cc" ;
// Function Call minimumCost(A, B); // This code is contributed by importantly. </script> |
7
Time Complexity: O(N + M)
Auxiliary Space: O(1) because constant size arrays fre1 and fre2 are used