Minimize the String Transformation Cost

Given a string Str and cost array C[] of size n replacing Str[i] with any character will have cost C[i]. The task is to minimize the cost where every odd i, for all i from 1 to n have to match either,

• S[i] = S[n-i+1] and S[i+1] = S[n-i] or,
• S[i] = S[n-i] and S[i+1] = S[n-i+1]

Examples:

Input: N = 4, Str = bdbd, C[] = {1, 2, 3, 4}
Output: 0
Explanation: Given string is already 2-drome as all elements of it satisfy the second condition.

Input: N = 8, Str = abbacaaa, C[] = {1, 2, 3, 4, 2, 3, 0, 1}
Output: 2
Explanation: Replacing the 5th character in string Str with b and the 7th character with b will be optimal

Approach: This can be solved with the following idea:

Start iterating from starting and last position as i and j. Check for each option given, if any option matched continue, or if not matched go for cost but select minimum from the options given.

Below are the steps involved:

• Iterate from starting as i and from last as j.
• if given option got matched, Continue the process
• Otherwise, go for:
  • firstCost = min(c[i], c[j]) + min(c[i + 1], c[j – 1])
  • secondCost = min(c[i], c[j – 1]) + min(c[i + 1], c[j])
• Choose minimum from them, add it to result.

Below is the implementation of the above idea:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function to find minimum cost
int solve(int n, string Str, int c[])
{
    int cost = 0;

    // Iterate from first and last
    for (int i = 0, j = n - 1; i < j; i += 2, j -= 2) {
        if (Str[i] == Str[j] && Str[i + 1] == Str[j - 1])
            continue;

        if (Str[i] == Str[j - 1] && Str[i + 1] == Str[j])
            continue;

        // 1st condition
        int firstCost = 0, secondCost = 0;

        // Checking the cost minimum for each
        // and every option
        if (Str[i] != Str[j])
            firstCost += min(c[i], c[j]);

        if (Str[i + 1] != Str[j - 1])
            firstCost += min(c[i + 1], c[j - 1]);

        if (Str[i] != Str[j - 1])
            secondCost += min(c[i], c[j - 1]);

        if (Str[i + 1] != Str[j])
            secondCost += min(c[i + 1], c[j]);

        // Add minimum cost
        cost += min(firstCost, secondCost);
    }

    // Return the cost
    return cost;
}

// Driver code
int main()
{
    int n = 4;
    string Str = "bdbd";
    int c[] = { 1, 2, 3, 4 };

    // Function call
    cout << solve(n, Str, c);

    return 0;
}

Java

// Java code for the above approach:
import java.util.*;

public class Main {

    // Function to find the minimum cost
    public static int solve(int n, String Str, int[] c)
    {
        int cost = 0;

        // Iterate from first and last
        for (int i = 0, j = n - 1; i < j; i += 2, j -= 2) {
            if (Str.charAt(i) == Str.charAt(j)
                && Str.charAt(i + 1) == Str.charAt(j - 1))
                continue;

            if (Str.charAt(i) == Str.charAt(j - 1)
                && Str.charAt(i + 1) == Str.charAt(j))
                continue;

            // 1st condition
            int firstCost = 0, secondCost = 0;

            // Checking the minimum cost for each and every
            // option
            if (Str.charAt(i) != Str.charAt(j))
                firstCost += Math.min(c[i], c[j]);

            if (Str.charAt(i + 1) != Str.charAt(j - 1))
                firstCost += Math.min(c[i + 1], c[j - 1]);

            if (Str.charAt(i) != Str.charAt(j - 1))
                secondCost += Math.min(c[i], c[j - 1]);

            if (Str.charAt(i + 1) != Str.charAt(j))
                secondCost += Math.min(c[i + 1], c[j]);

            // Add the minimum cost
            cost += Math.min(firstCost, secondCost);
        }

        // Return the cost
        return cost;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 4;
        String Str = "bdbd";
        int[] c = { 1, 2, 3, 4 };

        // Function call
        System.out.println(solve(n, Str, c));
    }
}

Python

# code added by flutterfly
def solve(n, Str, c):
    cost = 0
    # Iterate from first and last
    for i in range(0, n - 1, 2):
        j = n - 1 - i
        if Str[i] == Str[j] and Str[i + 1] == Str[j - 1]:
            continue
        if Str[i] == Str[j - 1] and Str[i + 1] == Str[j]:
            continue
        # 1st condition
        firstCost = 0
        secondCost = 0
        # Checking the cost minimum for each
        # and every option
        if Str[i] != Str[j]:
            firstCost += min(c[i], c[j])
        if Str[i + 1] != Str[j - 1]:
            firstCost += min(c[i + 1], c[j - 1])
        if Str[i] != Str[j - 1]:
            secondCost += min(c[i], c[j - 1])
        if Str[i + 1] != Str[j]:
            secondCost += min(c[i + 1], c[j])
        # Add minimum cost
        cost += min(firstCost, secondCost)
    # Add minimum cost
    return cost
# Driver code
n = 4
Str = "bdbd"
c = [1, 2, 3, 4]
# Function's call
print(solve(n, Str, c))

C#

using System;

class Program
{
  // Function to find minimum cost
    static int Solve(int n, string Str, int[] c)
    {
        int cost = 0;
        // Iterate from first and last
        for (int i = 0, j = n - 1; i < j; i += 2, j -= 2)
        {
            if (Str[i] == Str[j] && Str[i + 1] == Str[j - 1])
                continue;
            if (Str[i] == Str[j - 1] && Str[i + 1] == Str[j])
                continue;
            // 1st condition
            int firstCost = 0, secondCost = 0;
            // Checking the cost minimum for each
            // and every option
            if (Str[i] != Str[j])
                firstCost += Math.Min(c[i], c[j]);
            if (Str[i + 1] != Str[j - 1])
                firstCost += Math.Min(c[i + 1], c[j - 1]);
            if (Str[i] != Str[j - 1])
                secondCost += Math.Min(c[i], c[j - 1]);
            if (Str[i + 1] != Str[j])
                secondCost += Math.Min(c[i + 1], c[j]);
            // Add minimum cost
            cost += Math.Min(firstCost, secondCost);
        }
        // Return the cost
        return cost;
    }
// Driver code
    static void Main()
    {
        int n = 4;
        string Str = "bdbd";
        int[] c = { 1, 2, 3, 4 };
        // Function's Call
        Console.WriteLine(Solve(n, Str, c));
    }
}

Javascript

// code added by Flutterfly

// Function to find minimum cost
function solve(n, Str, c) {
    let cost = 0;
    // Iterate from first and last
    for (let i = 0, j = n - 1; i < j; i += 2, j -= 2) {
        if (Str[i] == Str[j] && Str[i + 1] == Str[j - 1])
            continue;
        if (Str[i] == Str[j - 1] && Str[i + 1] == Str[j])
            continue;
        // 1st condition
        let firstCost = 0, secondCost = 0;
        // Checking the cost minimum for each
        // and every option
        if (Str[i] != Str[j])
            firstCost += Math.min(c[i], c[j]);
        if (Str[i + 1] != Str[j - 1])
            firstCost += Math.min(c[i + 1], c[j - 1]);
        if (Str[i] != Str[j - 1])
            secondCost += Math.min(c[i], c[j - 1]);
        if (Str[i + 1] != Str[j])
            secondCost += Math.min(c[i + 1], c[j]);
        // Add minimum cost
        cost += Math.min(firstCost, secondCost);
    }
    // Return the cost
    return cost;
}
// Driver code
const n = 4;
const Str = "bdbd";
const c = [1, 2, 3, 4];
// Function's Call
console.log(solve(n, Str, c));

Output

0

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