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# Minimum Cost to make two Numeric Strings Identical

Given two numeric strings, A and B. A numeric string is a string that contains only digits [‘0’-‘9’].
The task is to make both the strings equal in minimum cost. The only operation that you are allowed to do is to delete a character (i.e. digit) from any of the strings (A or B). The cost of deleting a digit D is D units.

Examples

Input : A = “7135”, B = “135”
Output : 7
To make both string identical we have to delete ‘7’ from string A.

Input : A = “9142”, B = “1429”
Output : 14
There are 2 ways to make string “9142” identical to “1429” i.e either by deleting ‘9’ from both the strings or by deleting ‘1’, ‘4’and ‘2’ from both the string. Deleting 142 from both the string will cost 2*(1+4+2)=14 which is more optimal than deleting ‘9’.

This problem is a variation of a popular Dynamic Programming problem – Longest Common Subsequence. The idea is to find the maximum weight common subsequence which will be our required optimal identical string. To find the cost of deletion, subtract the sum of maximum weight common subsequence from the sum of string A and B.

Minimum weight to make string identical = costA + costB – 2*(cost of LCS)

Implementation:

## C++

 `// CPP program to find minimum cost to make` `// two numeric strings identical`   `#include `   `using` `namespace` `std;`   `typedef` `long` `long` `int` `ll;`   `// Function to find weight of LCS` `int` `lcs(``int` `dp, string a, string b,` `        ``int` `m, ``int` `n)` `{` `    ``for` `(``int` `i = 0; i < 100; i++)` `        ``for` `(``int` `j = 0; j < 100; j++)` `            ``dp[i][j] = -1;`   `    ``if` `(m < 0 || n < 0) {` `        ``return` `0;` `    ``}`   `    ``// if this state is already` `    ``// calculated then return` `    ``if` `(dp[m][n] != -1)` `        ``return` `dp[m][n];`   `    ``int` `ans = 0;` `    ``if` `(a[m] == b[n]) {` `        ``// adding required weight for` `        ``// particular match` `        ``ans = ``int``(a[m] - 48) + lcs(dp, a, b,` `                                   ``m - 1, n - 1);` `    ``}` `    ``else` `        ``// recurse for left and right child` `        ``// and store the max` `        ``ans = max(lcs(dp, a, b, m - 1, n),` `                  ``lcs(dp, a, b, m, n - 1));`   `    ``dp[m][n] = ans;` `    ``return` `ans;` `}`   `// Function to calculate cost of string` `int` `costOfString(string str)` `{` `    ``int` `cost = 0;`   `    ``for` `(``int` `i = 0; i < str.length(); i++)` `        ``cost += ``int``(str[i] - 48);`   `    ``return` `cost;` `}`   `// Driver code` `int` `main()` `{` `    ``string a, b;`   `    ``a = ``"9142"``;` `    ``b = ``"1429"``;`   `    ``int` `dp;`   `    ``// Minimum cost needed to make two strings identical` `    ``cout << (costOfString(a) + costOfString(b) - ` `                       ``2 * lcs(dp, a, b, a.length() - 1, ` `                                       ``b.length() - 1));`   `    ``return` `0;` `}`

## Java

 `// Java program to find minimum cost to make` `// two numeric strings identical`   `import` `java.io.*;`   `class` `GFG {` ` `    `// Function to find weight of LCS` ` ``static` `int` `lcs(``int` `dp[][], String a, String b,` `        ``int` `m, ``int` `n)` `{` `    ``for` `(``int` `i = ``0``; i < ``100``; i++)` `        ``for` `(``int` `j = ``0``; j < ``100``; j++)` `            ``dp[i][j] = -``1``;`   `    ``if` `(m < ``0` `|| n < ``0``) {` `        ``return` `0``;` `    ``}`   `    ``// if this state is already` `    ``// calculated then return` `    ``if` `(dp[m][n] != -``1``)` `        ``return` `dp[m][n];`   `    ``int` `ans = ``0``;` `    ``if` `(a.charAt(m) == b.charAt(n)) {` `        ``// adding required weight for` `        ``// particular match` `        ``ans = (a.charAt(m) - ``48``) + lcs(dp, a, b,` `                                ``m - ``1``, n - ``1``);` `    ``}` `    ``else` `        ``// recurse for left and right child` `        ``// and store the max` `        ``ans = Math.max(lcs(dp, a, b, m - ``1``, n),` `                ``lcs(dp, a, b, m, n - ``1``));`   `    ``dp[m][n] = ans;` `    ``return` `ans;` `}`   `// Function to calculate cost of string` ` ``static` `int` `costOfString(String str)` `{` `    ``int` `cost = ``0``;`   `    ``for` `(``int` `i = ``0``; i < str.length(); i++)` `        ``cost += (str.charAt(i) - ``48``);`   `    ``return` `cost;` `}`   `// Driver code` `    ``public` `static` `void` `main (String[] args) {` `            ``String a, b;`   `    ``a = ``"9142"``;` `    ``b = ``"1429"``;`   `    ``int` `dp[][] = ``new` `int``[``101``][``101``];`   `    ``// Minimum cost needed to make two strings identical` `    ``System.out.print( (costOfString(a) + costOfString(b) - ` `                    ``2` `* lcs(dp, a, b, a.length() - ``1``, ` `                                    ``b.length() - ``1``)));`   `    ``}` `}` `// This code is contributed by anuj_67.`

## Python 3

 `# Python 3 program to find minimum cost ` `# to make two numeric strings identical`   `# Function to find weight of LCS` `def` `lcs(dp, a, b, m, n):`   `    ``for` `i ``in` `range``(``100``):` `        ``for` `j ``in` `range``(``100``):` `            ``dp[i][j] ``=` `-``1`   `    ``if` `(m < ``0` `or` `n < ``0``) :` `        ``return` `0`   `    ``# if this state is already calculated ` `    ``# then return` `    ``if` `(dp[m][n] !``=` `-``1``):` `        ``return` `dp[m][n]`   `    ``ans ``=` `0` `    ``if` `(a[m] ``=``=` `b[n]):` `        `  `        ``# adding required weight for` `        ``# particular match` `        ``ans ``=` `(``ord``(a[m]) ``-` `48``) ``+` `lcs(dp, a, b,` `                                     ``m ``-` `1``, n ``-` `1``)` `    `  `    ``else``:` `        `  `        ``# recurse for left and right child` `        ``# and store the max` `        ``ans ``=` `max``(lcs(dp, a, b, m ``-` `1``, n),` `                  ``lcs(dp, a, b, m, n ``-` `1``))`   `    ``dp[m][n] ``=` `ans` `    ``return` `ans`   `# Function to calculate cost of string` `def` `costOfString(s):` `    `  `    ``cost ``=` `0`   `    ``for` `i ``in` `range``(``len``(s)):` `        ``cost ``+``=` `(``ord``(s[i]) ``-` `48``)`   `    ``return` `cost`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``a ``=` `"9142"` `    ``b ``=` `"1429"`   `    ``dp ``=` `[[``0` `for` `x ``in` `range``(``101``)] ` `             ``for` `y ``in` `range``(``101``)]`   `    ``# Minimum cost needed to make two` `    ``# strings identical` `    ``print``(costOfString(a) ``+` `costOfString(b) ``-` `2` `*` `           ``lcs(dp, a, b, ``len``(a) ``-` `1``, ``len``(b) ``-` `1``))`   `# This code is contributed by ita_c`

## C#

 `// C# program to find minimum cost to make ` `// two numeric strings identical ` `using` `System;` `public` `class` `GFG { `   `// Function to find weight of LCS ` `static` `int` `lcs(``int` `[,]dp, String a, String b, ` `        ``int` `m, ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < 100; i++) ` `        ``for` `(``int` `j = 0; j < 100; j++) ` `            ``dp[i,j] = -1; `   `    ``if` `(m < 0 || n < 0) { ` `        ``return` `0; ` `    ``} `   `    ``// if this state is already ` `    ``// calculated then return ` `    ``if` `(dp[m,n] != -1) ` `        ``return` `dp[m,n]; `   `    ``int` `ans = 0; ` `    ``if` `(a[m] == b[n]) { ` `        ``// adding required weight for ` `        ``// particular match ` `        ``ans = (a[m] - 48) + lcs(dp, a, b, m - 1, n - 1); ` `    ``} ` `    ``else` `        ``// recurse for left and right child ` `        ``// and store the max ` `        ``ans = Math.Max(lcs(dp, a, b, m - 1, n), ` `                ``lcs(dp, a, b, m, n - 1)); `   `    ``dp[m,n] = ans; ` `    ``return` `ans; ` `} `   `// Function to calculate cost of string ` `static` `int` `costOfString(String str) ` `{ ` `    ``int` `cost = 0; `   `    ``for` `(``int` `i = 0; i < str.Length; i++) ` `        ``cost += (str[i] - 48); `   `    ``return` `cost; ` `} `   `// Driver code ` `    ``public` `static` `void` `Main () { ` `            ``String a, b; `   `    ``a = ``"9142"``; ` `    ``b = ``"1429"``; `   `    ``int` `[,]dp = ``new` `int``[101,101]; `   `    ``// Minimum cost needed to make two strings identical ` `    ``Console.Write( (costOfString(a) + costOfString(b) - ` `                ``2 * lcs(dp, a, b, a.Length- 1, ` `                ``b.Length - 1))); `   `    ``} ` `} ` `// This code is contributed by Rajput-Ji `

## Javascript

 ``

Output

`14`

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