Minimum Cost To Make Two Strings Identical

Given two strings X and Y, and two values costX and costY. We need to find minimum cost required to make the given two strings identical. We can delete characters from both the strings. The cost of deleting a character from string X is costX and from Y is costY. Cost of removing all characters from a string is same.

Examples :

Input :  X = "abcd", Y = "acdb", costX = 10, costY = 20.
Output:  30
For Making both strings identical we have to delete 
character 'b' from both the string, hence cost will
be = 10 + 20 = 30.

Input :  X = "ef", Y = "gh", costX = 10, costY = 20.
Output:  60
For making both strings identical, we have to delete 2-2
characters from both the strings, hence cost will be =
 10 + 10 + 20 + 20 = 60.



This problem is a variation of Longest Common Subsequence ( LCS ). The idea is simple, we first find the length of longest common subsequence of strings X and Y. Now subtracting len_LCS with lengths of individual strings gives us number of characters to be removed to make them identical.

// Cost of making two strings identical is SUM of following two
//   1) Cost of removing extra characters (other than LCS) 
//      from X[]
//   2) Cost of removing extra characters (other than LCS) 
//      from Y[]
Minimum Cost to make strings identical = costX * (m - len_LCS) + 
                                         costY * (n - len_LCS).  

m ==> Length of string X
m ==> Length of string Y
len_LCS ==> Length of LCS Of X and Y.
costX ==> Cost of removing a character from X[]
costY ==> Cost of removing a character from Y[]

Note that cost of removing all characters from a string
is same.               

Below is the implementation of above idea.

C++

/* C++ code to find minimum cost to make two strings
   identical */
#include<bits/stdc++.h>
using namespace std;
  
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs(char *X, char *Y, int m, int n)
{
    int L[m+1][n+1];
  
    /* Following steps build L[m+1][n+1] in bottom
       up fashion. Note that L[i][j] contains length
       of LCS of X[0..i-1] and Y[0..j-1] */
    for (int i=0; i<=m; i++)
    {
        for (int j=0; j<=n; j++)
        {
            if (i == 0 || j == 0)
                L[i][j] = 0;
  
            else if (X[i-1] == Y[j-1])
                L[i][j] = L[i-1][j-1] + 1;
  
            else
                L[i][j] = max(L[i-1][j], L[i][j-1]);
        }
    }
  
    /* L[m][n] contains length of LCS for X[0..n-1] and
       Y[0..m-1] */
    return L[m][n];
}
  
// Returns cost of making X[] and Y[] identical.  costX is
// cost of removing a character from X[] and costY is cost
// of removing a character from Y[]/
int findMinCost(char X[], char Y[], int costX, int costY)
{
    // Find LCS of X[] and Y[]
    int m = strlen(X), n = strlen(Y);
    int len_LCS = lcs(X, Y, m, n);
  
    // Cost of making two strings identical is SUM of
    // following two
    //   1) Cost of removing extra characters
    //      from first string
    //   2) Cost of removing extra characters from
    //      second string
    return costX * (m - len_LCS) +
           costY * (n - len_LCS);
}
  
/* Driver program to test above function */
int main()
{
    char X[] = "ef";
    char Y[] = "gh";
    cout << "Minimum Cost to make two strings "
         << " identical is = " << findMinCost(X, Y, 10, 20);
    return 0;
}

Java

// Java code to find minimum cost to
// make two strings identical 
import java.io.*;
  
class GFG {
      
    // Returns length of LCS for X[0..m-1], Y[0..n-1] 
    static int lcs(String X, String Y, int m, int n)
    {
        int L[][]=new int[m + 1][n + 1];
      
        /* Following steps build L[m+1][n+1] in bottom
        up fashion. Note that L[i][j] contains length
        of LCS of X[0..i-1] and Y[0..j-1] */
        for (int i = 0; i <= m; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                if (i == 0 || j == 0)
                    L[i][j] = 0;
      
                else if (X.charAt(i - 1) == Y.charAt(j - 1))
                    L[i][j] = L[i - 1][j - 1] + 1;
      
                else
                    L[i][j] = Math.max(L[i - 1][j], L[i][j - 1]);
            }
        }
      
        // L[m][n] contains length of LCS 
        // for X[0..n-1] and Y[0..m-1] 
        return L[m][n];
    }
      
    // Returns cost of making X[] and Y[] identical. 
    // costX is cost of removing a character from X[] 
    // and costY is cost of removing a character from Y[]/
    static int findMinCost(String X, String Y, int costX, int costY)
    {
        // Find LCS of X[] and Y[]
        int m = X.length();
        int n = Y.length();
        int len_LCS;
        len_LCS = lcs(X, Y, m, n);
      
        // Cost of making two strings identical
        //  is SUM of following two
        // 1) Cost of removing extra characters
        // from first string
        // 2) Cost of removing extra characters
        // from second string
        return costX * (m - len_LCS) +
               costY * (n - len_LCS);
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        String X = "ef";
        String Y = "gh";
        System.out.println( "Minimum Cost to make two strings "
                            + " identical is = " 
                            + findMinCost(X, Y, 10, 20));
          
    }
}
  
// This code is contributed by vt_m

Python3

# Python code to find minimum cost 
# to make two strings identical
  
# Returns length of LCS for
# X[0..m-1], Y[0..n-1] 
def lcs(X, Y, m, n):
    L = [[0 for i in range(n + 1)] 
            for i in range(m + 1)] 
              
    # Following steps build 
    # L[m+1][n+1] in bottom 
    # up fashion. Note that 
    # L[i][j] contains length 
    # of LCS of X[0..i-1] and Y[0..j-1]
    for i in range(m + 1):
        for j in range(n + 1):
            if i == 0 or j == 0:
                L[i][j] = 0
            elif X[i - 1] == Y[j - 1]:
                L[i][j] = L[i - 1][j - 1] + 1
            else:
                L[i][j] = max(L[i - 1][j], 
                              L[i][j - 1])
        # L[m][n] contains length of 
        # LCS for X[0..n-1] and Y[0..m-1]
    return L[m][n]
      
# Returns cost of making X[] 
# and Y[] identical. costX is 
# cost of removing a character
# from X[] and costY is cost 
# of removing a character from Y[]
def findMinCost(X, Y, costX, costY):
      
    # Find LCS of X[] and Y[]
    m = len(X)
    n = len(Y)
    len_LCS =lcs(X, Y, m, n)
      
    # Cost of making two strings 
    # identical is SUM of following two 
    # 1) Cost of removing extra  
    # characters from first string 
    # 2) Cost of removing extra 
    # characters from second string
    return (costX * (m - len_LCS) +
            costY * (n - len_LCS))
              
# Driver Code
X = "ef"
Y = "gh"
print('Minimum Cost to make two strings ', end = '')
print('identical is = ', findMinCost(X, Y, 10, 20))
  
# This code is contributed
# by sahilshelangia

C#

// C# code to find minimum cost to
// make two strings identical 
using System; 
  
class GFG {
      
    // Returns length of LCS for X[0..m-1], Y[0..n-1] 
    static int lcs(String X, String Y, int m, int n)
    {
        int [,]L = new int[m + 1, n + 1];
      
        /* Following steps build L[m+1][n+1] in bottom
           up fashion. Note that L[i][j] contains length
           of LCS of X[0..i-1] and Y[0..j-1] */
        for (int i = 0; i <= m; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                if (i == 0 || j == 0)
                    L[i,j] = 0;
      
                else if (X[i - 1] == Y[j - 1])
                    L[i,j] = L[i - 1,j - 1] + 1;
      
                else
                    L[i,j] = Math.Max(L[i - 1,j], L[i,j - 1]);
            }
        }
      
        // L[m][n] contains length of LCS 
        // for X[0..n-1] and Y[0..m-1] 
        return L[m,n];
    }
      
    // Returns cost of making X[] and Y[] identical.
    // costX is cost of removing a character from X[] 
    // and costY is cost of removing a character from Y[]
    static int findMinCost(String X, String Y, 
                           int costX, int costY)
    {
        // Find LCS of X[] and Y[]
        int m = X.Length;
        int n = Y.Length;
        int len_LCS;
        len_LCS = lcs(X, Y, m, n);
      
        // Cost of making two strings identical
        // is SUM of following two
        // 1) Cost of removing extra characters
        // from first string
        // 2) Cost of removing extra characters
        // from second string
        return costX * (m - len_LCS) +
               costY * (n - len_LCS);
    }
      
    // Driver code
    public static void Main () 
    {
        String X = "ef";
        String Y = "gh";
        Console.Write( "Minimum Cost to make two strings " +
                       " identical is = " +
                       findMinCost(X, Y, 10, 20));
    }
}
  
// This code is contributed by nitin mittal.


Output:



  Minimum Cost to make two strings identical is = 60

This article is contributed by Shashank Mishra ( Gullu ). This article is reviwed by team geeksforgeeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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