Minimum cost to convert one given string to another using swap, insert or delete operations
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++
#include <bits/stdc++.h>
using namespace std;
void minimumCost(string a, string b)
{
vector< int > fre1(256), fre2(256);
for ( char c : a)
fre1[( int )(c)]++;
for ( char c : b)
fre2[( int )(c)]++;
int mincost = 0;
for ( int i = 0; i < 256; i++) {
mincost += abs (fre1[i]
- fre2[i]);
}
cout << mincost << endl;
}
int main()
{
string A = "1AB+-" , B = "cc" ;
minimumCost(A, B);
return 0;
}
|
Java
import java.util.*;
class GFG{
public static void minimumCost(String a, String b)
{
int fre1[] = new int [ 256 ];
int fre2[] = new int [ 256 ];
for ( char c : a.toCharArray())
fre1[( int )(c)]++;
for ( char c : b.toCharArray())
fre2[( int )(c)]++;
int mincost = 0 ;
for ( int i = 0 ; i < 256 ; i++)
{
mincost += Math.abs(fre1[i] -
fre2[i]);
}
System.out.println(mincost);
}
public static void main(String[] args)
{
String A = "1AB+-" , B = "cc" ;
minimumCost(A, B);
}
}
|
Python3
def minimumCost(a, b):
fre1 = [ 0 ] * ( 256 )
fre2 = [ 0 ] * ( 256 )
for c in a:
fre1[ ord (c)] + = 1
for c in b:
fre2[ ord (c)] + = 1
mincost = 0
for i in range ( 256 ):
mincost + = abs (fre1[i] - fre2[i])
print (mincost)
if __name__ = = '__main__' :
A = "1AB+-"
B = "cc"
minimumCost(A, B)
|
C#
using System;
using System.Collections.Generic;
class GFG{
public static void minimumCost( string a,
string b)
{
int [] fre1 = new int [256];
int [] fre2 = new int [256];
foreach ( char c in a.ToCharArray())
fre1[( int )(c)]++;
foreach ( char c in b.ToCharArray())
fre2[( int )(c)]++;
int mincost = 0;
for ( int i = 0; i < 256; i++)
{
mincost += Math.Abs(fre1[i] -
fre2[i]);
}
Console.Write(mincost);
}
public static void Main()
{
string A = "1AB+-" , B = "cc" ;
minimumCost(A, B);
}
}
|
Javascript
<script>
function minimumCost(a, b)
{
var fre1 = Array(256).fill(0), fre2= Array(256).fill(0);
a.split( '' ).forEach(c => {
fre1[c.charCodeAt(0)]++;
});
b.split( '' ).forEach(c => {
fre2[c.charCodeAt(0)]++;
});
var mincost = 0;
for ( var i = 0; i < 256; i++) {
mincost += Math.abs(fre1[i]
- fre2[i]);
}
document.write( mincost );
}
var A = "1AB+-" , B = "cc" ;
minimumCost(A, B);
</script>
|
Time Complexity: O(N + M)
Auxiliary Space: O(1) because constant size arrays fre1 and fre2 are used
Last Updated :
11 Nov, 2022
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