The problem can be solved recursively,

Let,

    N = length of string1
    M = length of string1
    F(n,m) = minimum cost to make two strings similar

Let us consider, what can be cases that can arrive for F(i,j) where 0 <= i<n && 0 <= j<m

Case 1.

string1[i]==string2[j] that is indexed characters are similar

In such a case, we don't need any additional cost to make strings similar, the cost will be similar to sub-problem of size i-1, j-1. This can be recursively written as

    F(i,j) = F(i-1,j-1) if string1[i] == string2[j]

Case 2.

string1[i]!=string2[j] that is indexed characters are not similar.

In such a case we need to invest,

1. One thing can remove the indexed character from string1 and change to indexed string2 character.
   This can be recursively written as,

       F(i,j) = F(i-1,j) + costX if string1[i] != string2[j]
   

2. Another way can be to remove the indexed character from string2 and change to string1 indexed character.
   This can be recursively written as,

       F(i,j) = F(i,j-1) + costY if string1[i] != string2[j]
   

3. Remove characters from both.
   This can be recursively written as,

       F(i,j) = F(i-1,j-1) + costY + costX if string1[i] != string2[j]
   

   Finally, we would take minimum out of this three cases.

So here goes the problem structure,

Now the above recursion will create many overlapping subproblems and hence we need two converts it into DP.

Converting into DP

    n = string1 length
    m = string2 length
    str1 = string1
    str2 = string2

    1)  Declare int dp[n+1][m+1];
    2)  Base case
            for i=0 to n
                dp[i][0]=i*costX;
            for j=0 to m
                dp[0][j]=j*costY;
    3)  Fill the DP
            for i=1 to n
                for j=1 to m
                    //first case when str1[i-1]==str2[j-1]
                    dp[i][j]=dp[i-1][j-1];  
                    if(str1[i-1]!=str2[j-1])
                        dp[i][j]+=costX+costY;
                        dp[i][j]=min(dp[i][j],min(dp[i-1][j]+costX,dp[i][j-1]+costY));
                    end if
            
                end for
            end for
    4)  return dp[n][m]; 

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

int editdistance(string s1, string s2, int costX, int costY)
{

    int n = s1.length();
    int m = s2.length();
    int dp[n + 1][m + 1];
    
    for (int i = 0; i <= n; i++)
        dp[i][0] = i * costX;
    
    for (int j = 0; j <= m; j++)
        dp[0][j] = j * costY;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            dp[i][j] = dp[i - 1][j - 1];
            if (s1[i - 1] != s2[j - 1]) {
                dp[i][j] += costX + costY;
                dp[i][j] = min(dp[i][j], min(dp[i - 1][j] + costX, dp[i][j - 1] + costY));
            }
        }
    }
    
    return dp[n][m];
}

int main()
{
    int costX, costY;
    
    cout << "Cost of changing/removing character from string1: \n";
    cin >> costX;
    
    cout << "Cost of changing/removing character from string2: \n";
    cin >> costY;
    
    string s1, s2;
    
    cout << "Input string1\n";
    cin >> s1;
    cout << "Input string2\n";
    cin >> s2;
    
    cout << "Minimum cost to make two string similar: " << editdistance(s1, s2, costX, costY) << endl;

    return 0;
}

Output

Cost of changing/removing character from string1: 
2
Cost of changing/removing character from string2: 
3
Input string1
includehelp
Input string2
includuggu
Minimum cost to make two string similar: 22