Minimum cost to make the String palindrome using ASCII values

Given an input string S, the task is to make it a palindrome and calculate the minimum cost for making it a palindrome, where you are allowed to use any one of the following operations any number of times:

• For replacing the current character with another character the cost will be the absolute difference in ASCII (American Standard Code for Information Interchange) values of these two characters 
• Replace ‘a‘ with both characters and this will cost 10. If the input string is already palindrome then return -1.

Examples:

Input: S = “moon”
Output: 1
Explanation: We need to make ‘m’ = = ‘n’ then the input string will become palindrome. So, the cost will be the ASCII value of ‘n’ – ASCII value of ‘m’ i.e. 1.

Input: S = “zone”
Output: 11
Explanation: We need to make ‘z’ == ‘e’ and the absolute difference between their ASCII values is 21. So we replace both characters with ‘a’ and it will cost 10 which is more efficient than 21 and for making ‘o’ == ‘n’ the cost will be 1. Total cost = 10 + 1 .

Approach: This can be solved with the following idea:

For every pair of characters choose whether their ASCII difference gives an optimized answer or replacing them with ‘a’ gives the optimized answer.

Below are the steps involved in the implementation of the code:

• Initialize a variable minCost = 0 for tracking the cost.
• Iterate over the input string using a for loop from 0 to n/2(n is the size of the input string) and for every single iteration do the below-mentioned checks.
• If the absolute difference of ASCII values is smaller than 10 then add it to minCost otherwise add 10 to minCost.
• After the for loop is executed then check whether minCost is greater than 0 or not. If yes then return minCost otherwise return -1.

Below is the implementation of the above-mentioned approach in C++:

C++

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

// Function to find cost for making it
// palindrome
int findminCost(string s)
{

    // Initializing variable for
    // calculating minimum cost
    int minCost = 0;
    int n = s.size();

    for (int i = 0; i < n / 2; i++) {
        if (s[i] != s[n - i - 1]) {

            // If difference of ascii is
            // greater than 10 then we
            // replace both character
            // with 'a' and add 10 to
            // minCost
            minCost += min(10, abs(s[i] - s[n - i - 1]));
        }
    }
    if (minCost > 0)
        return minCost;

    // Returning -1 if the string is
    // already palindrome
    return -1;
}

// Driver code
int main()
{
    string s = "moon";

    // Function call
    cout << findminCost(s);

    return 0;
}

Java

// C++ Implementation
public class MinimumCostPalindrome {
    // Function to find cost for making it
    // palindrome
    static int findMinCost(String s) {
        // Initializing variable for
        // calculating minimum cost
        int minCost = 0;
        int n = s.length();

        for (int i = 0; i < n / 2; i++) {
            if (s.charAt(i) != s.charAt(n - i - 1)) {
                // If difference of ASCII is
                // greater than 10 then we
                // replace both characters
                // with 'a' and add 10 to
                // minCost
                minCost += Math.min(10, Math.abs(s.charAt(i) - s.charAt(n - i - 1)));
            }
        }

        if (minCost > 0)
            return minCost;

        // Returning -1 if the string is
        // already palindrome
        return -1;
    }

    // Driver code
    public static void main(String[] args) {
        String s = "moon";

        // Function call
        System.out.println(findMinCost(s));
    }
}
// This code is contributed by codearcade

Python

# Function to find cost for making it
# palindrome

def find_min_cost(s):

    # Initializing variable for
    # calculating minimum cost
    min_cost = 0
    n = len(s)

    for i in range(n // 2):
        if s[i] != s[n - i - 1]:

            # If difference of ASCII is
            # greater than 10, then we
            # replace both characters
            # with 'a' and add 10 to
            # min_cost
            min_cost += min(10, abs(ord(s[i]) - ord(s[n - i - 1])))

    if min_cost > 0:
        return min_cost

    # Returning -1 if the string is
    # already a palindrome
    return -1

# Driver code


def main():
    s = "moon"

    # Function call
    print(find_min_cost(s))


if __name__ == "__main__":
    main()

C#

// C# code  of the above approach
using System;

public class MinimumCostPalindrome
{
    // Function to find cost for making it
    // palindrome
    static int FindMinCost(string s)
    {
        // Initializing variable for
        // calculating minimum cost
        int minCost = 0;
        int n = s.Length;

        for (int i = 0; i < n / 2; i++)
        {
            if (s[i] != s[n - i - 1])
            {
                // If difference of ASCII is
                // greater than 10 then we
                // replace both characters
                // with 'a' and add 10 to
                // minCost
                minCost += Math.Min(10, Math.Abs(s[i] - s[n - i - 1]));
            }
        }

        if (minCost > 0)
            return minCost;

        // Returning -1 if the string is
        // already palindrome
        return -1;
    }

    // Driver code
    public static void Main(string[] args)
    {
        string s = "moon";

        // Function call
        Console.WriteLine(FindMinCost(s));
    }
}

// This code is contributed by Dwaipayan Bandyopadhyay

Javascript

// Nikunj Sonigara

// Function to find cost for making it palindrome
function findminCost(s) {
    // Initializing variable for calculating minimum cost
    let minCost = 0;
    const n = s.length;

    for (let i = 0; i < Math.floor(n / 2); i++) {
        if (s[i] !== s[n - i - 1]) {
            // If difference of ASCII is greater than 10, then we
            // replace both characters with 'a' and add 10 to minCost
            minCost += Math.min(10, Math.abs(s.charCodeAt(i) - s.charCodeAt(n - i - 1)));
        }
    }
    if (minCost > 0) {
        return minCost;
    }

    // Returning -1 if the string is already a palindrome
    return -1;
}

// Driver code
const s = "moon";
// Function call
console.log(findminCost(s));

Output

1

Time Complexity: O(N) 
Auxiliary Space: O(1)