# 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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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. ` |

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## PHP

`<?php ` `/* PHP code to find minimum cost to make two strings ` ` ` `identical */` ` ` `/* Returns length of LCS for X[0..m-1], Y[0..n-1] */` `function` `lcs(` `$X` `, ` `$Y` `, ` `$m` `, ` `$n` `) ` `{ ` ` ` `$L` `= ` `array_fill` `(0,(` `$m` `+1),` `array_fill` `(0,(` `$n` `+1),NULL)); ` ` ` ` ` `/* Following steps build L[m+1][n+1] in bottom ` ` ` `up fashion. Note that L[i][j] contains length ` ` ` `of LCS of X[0..i-1] and Y[0..j-1] */` ` ` `for` `(` `$i` `=0; ` `$i` `<=` `$m` `; ` `$i` `++) ` ` ` `{ ` ` ` `for` `(` `$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[]/ ` `function` `findMinCost(&` `$X` `, &` `$Y` `,` `$costX` `, ` `$costY` `) ` `{ ` ` ` `// Find LCS of X[] and Y[] ` ` ` `$m` `= ` `strlen` `(` `$X` `); ` ` ` `$n` `= ` `strlen` `(` `$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` `- ` `$len_LCS` `); ` `} ` ` ` `/* Driver program to test above function */` `$X` `= ` `"ef"` `; ` `$Y` `= ` `"gh"` `; ` `echo` `"Minimum Cost to make two strings "` `. ` ` ` `" identical is = "` `. findMinCost(` `$X` `, ` `$Y` `, 10, 20); ` `return` `0; ` `?> ` |

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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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