# Minimum cost to make two strings same

Given four integers a, b, c, d and two strings S1 and S2 of equal length, which consist only of the characters ‘2’, ‘1’ and ‘0’.

1. Converting ‘1’ to ‘2’ or vice versa costs a.
2. Converting ‘2’ to ‘3’ or vice versa costs b.
3. Converting ‘3’ to ‘1’ or vice versa costs c.
4. Deleting the ith character from both the strings costs d.

The task is to find the minimum cost to make both the strings equal in configuration.

Examples:

Input: s1 = “121”, s2 = “223”, a = 2, b = 3, c = 4, d = 10
Output: 6
Change the first character from ‘1’ to ‘2’ which costs 2.
Change the third character from ‘3’ to ‘1’ which costs 4.

Input: s1 = ‘222’, s2 = ‘111’, a = 20, b = 30, c = 40, d = 1
Output: 3
Delete all characters which costs 3 which is the minimum possible.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Iterate in the string, and if s1[i] = s2[i] then perform no operation.
• If the s1[i] = ‘1’ and s2[i] = ‘2’ then perform the minimum cost operation from the following:
1. Delete s1[i] and s2[i] which costs d.
2. Change ‘1’ to ‘2’ which costs a
3. Change ‘1’ to ‘3’ and then ‘3’ to ‘2’ which costs b + c.
• If the s1[i] = ‘2’ and s2[i] = ‘3’ then perform the minimum cost operation from the following:
1. Delete s1[i] and s2[i] which costs d.
2. Change ‘2’ to ‘3’ which costs b.
3. Change ‘2’ to ‘1’ and then ‘1’ to ‘3’ which costs a + c.
• If the s1[i] = ‘3’ and s2[i] = ‘1’ then perform the minimum cost operation from the following:
1. Delete s1[i] and s2[i] which costs d.
2. Change ‘3’ to ‘1’ which costs c.
3. Change ‘3’ to ‘2’ and then ‘2’ to ‘1’ which costs b + a.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum cost to make the ` `// configuration of both the strings same ` `int` `findCost(string s1, string s2, ` `             ``int` `a, ``int` `b, ``int` `c, ``int` `d, ``int` `n) ` `{ ` `    ``int` `cost = 0; ` ` `  `    ``// Iterate and find the cost ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(s1[i] == s2[i]) ` `            ``continue``; ` `        ``else` `{ ` ` `  `            ``// Find the minimum cost ` `            ``if` `((s1[i] == ``'1'` `&& s2[i] == ``'2'``) ` `                ``|| (s2[i] == ``'1'` `&& s1[i] == ``'2'``)) ` `                ``cost += min(d, min(a, b + c)); ` `            ``else` `if` `((s1[i] == ``'2'` `&& s2[i] == ``'3'``) ` `                     ``|| (s2[i] == ``'2'` `&& s1[i] == ``'3'``)) ` `                ``cost += min(d, min(b, a + c)); ` `            ``else` `if` `((s1[i] == ``'1'` `&& s2[i] == ``'3'``) ` `                     ``|| (s2[i] == ``'1'` `&& s1[i] == ``'3'``)) ` `                ``cost += min(d, min(c, a + b)); ` `        ``} ` `    ``} ` `    ``return` `cost; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string s1 = ``"121"``; ` `    ``string s2 = ``"223"``; ` `    ``int` `a = 2, b = 3, c = 4, d = 10; ` `    ``int` `n = s1.size(); ` `    ``cout << findCost(s1, s2, a, b, c, d, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `// Function to return the minimum cost to make the ` `// configuration of both the strings same ` `static` `int` `findCost(String s1, String s2, ` `                    ``int` `a, ``int` `b, ``int` `c,  ` `                    ``int` `d, ``int` `n) ` `{ ` `    ``int` `cost = ``0``; ` ` `  `    ``// Iterate and find the cost ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``if` `(s1.charAt(i) == s2.charAt(i)) ` `            ``continue``; ` `        ``else`  `        ``{ ` ` `  `            ``// Find the minimum cost ` `            ``if` `((s1.charAt(i) == ``'1'` `&& s2.charAt(i) == ``'2'``) ||  ` `                ``(s2.charAt(i) == ``'1'` `&& s1.charAt(i) == ``'2'``)) ` `                ``cost += Math.min(d, Math.min(a, b + c)); ` `            ``else` `if` `((s1.charAt(i) == ``'2'` `&& s2.charAt(i) == ``'3'``) ||  ` `                     ``(s2.charAt(i) == ``'2'` `&& s1.charAt(i) == ``'3'``)) ` `                ``cost += Math.min(d, Math.min(b, a + c)); ` `            ``else` `if` `((s1.charAt(i) == ``'1'` `&& s2.charAt(i) == ``'3'``) || ` `                     ``(s2.charAt(i) == ``'1'` `&& s1.charAt(i) == ``'3'``)) ` `                ``cost += Math.min(d, Math.min(c, a + b)); ` `        ``} ` `    ``} ` `    ``return` `cost; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String s1 = ``"121"``; ` `    ``String s2 = ``"223"``; ` `    ``int` `a = ``2``, b = ``3``, c = ``4``, d = ``10``; ` `    ``int` `n = s1.length(); ` `    ``System.out.println(findCost(s1, s2, a, b, c, d, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech. `

## Python3

 `# Python 3 implementation of the approach ` ` `  `# Function to return the minimum cost to make  ` `# the configuration of both the strings same ` `def` `findCost(s1, s2, a, b, c, d, n): ` `    ``cost ``=` `0` ` `  `    ``# Iterate and find the cost ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(s1[i] ``=``=` `s2[i]): ` `            ``continue` `        ``else``: ` `             `  `            ``# Find the minimum cost ` `            ``if` `((s1[i] ``=``=` `'1'` `and` `s2[i] ``=``=` `'2'``) ``or`  `                ``(s2[i] ``=``=` `'1'` `and` `s1[i] ``=``=` `'2'``)): ` `                ``cost ``+``=` `min``(d, ``min``(a, b ``+` `c)) ` `            ``elif` `((s1[i] ``=``=` `'2'` `and` `s2[i] ``=``=` `'3'``) ``or` `                  ``(s2[i] ``=``=` `'2'` `and` `s1[i] ``=``=` `'3'``)): ` `                ``cost ``+``=` `min``(d, ``min``(b, a ``+` `c)) ` `            ``elif` `((s1[i] ``=``=` `'1'` `and` `s2[i] ``=``=` `'3'``) ``or` `                  ``(s2[i] ``=``=` `'1'` `and` `s1[i] ``=``=` `'3'``)): ` `                ``cost ``+``=` `min``(d, ``min``(c, a ``+` `b)) ` `    ``return` `cost ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``s1 ``=` `"121"` `    ``s2 ``=` `"223"` `    ``a ``=` `2` `    ``b ``=` `3` `    ``c ``=` `4` `    ``d ``=` `10` `    ``n ``=` `len``(s1) ` `    ``print``(findCost(s1, s2, a, b, c, d, n)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the minimum cost to make the ` `// configuration of both the strings same ` `static` `int` `findCost(``string` `s1, ``string` `s2, ` `            ``int` `a, ``int` `b, ``int` `c, ``int` `d, ``int` `n) ` `{ ` `    ``int` `cost = 0; ` ` `  `    ``// Iterate and find the cost ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``if` `(s1[i] == s2[i]) ` `            ``continue``; ` `        ``else`  `        ``{ ` ` `  `            ``// Find the minimum cost ` `            ``if` `((s1[i] == ``'1'` `&& s2[i] == ``'2'``) ` `                ``|| (s2[i] == ``'1'` `&& s1[i] == ``'2'``)) ` `                ``cost +=Math.Min(d,Math.Min(a, b + c)); ` `            ``else` `if` `((s1[i] == ``'2'` `&& s2[i] == ``'3'``) ` `                    ``|| (s2[i] == ``'2'` `&& s1[i] == ``'3'``)) ` `                ``cost +=Math.Min(d,Math.Min(b, a + c)); ` `            ``else` `if` `((s1[i] == ``'1'` `&& s2[i] == ``'3'``) ` `                    ``|| (s2[i] == ``'1'` `&& s1[i] == ``'3'``)) ` `                ``cost +=Math.Min(d,Math.Min(c, a + b)); ` `        ``} ` `    ``} ` `    ``return` `cost; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``string` `s1 = ``"121"``; ` `    ``string` `s2 = ``"223"``; ` `    ``int` `a = 2, b = 3, c = 4, d = 10; ` `    ``int` `n = s1.Length; ` `    ``Console.WriteLine(findCost(s1, s2, a, b, c, d, n)); ` `} ` `} ` ` `  `// This Code is Contributed by Code_Mech. `

Output:

```6
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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