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: 

1. Swap 'a' with 'D'. Therefore, A = "DBca".
2. 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:

1. Initialize two arrays a[] and b[] of length 256 to store the frequencies of each character in the strings A and B respectively.
2. Initialize a variable, say minCost, to store the minimum cost.
3. Traverse over the range [0, 255] using the variable i and at each iteration, increment minCost by abs(a[i] - b[i]).
4. 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