The Love-Letter Mystery


Problem Statement :


James found a love letter that his friend Harry has written to his girlfriend. James is a prankster, so he decides to meddle with the letter. He changes all the words in the letter into palindromes.

To do this, he follows two rules:

1. He can only reduce the value of a letter by 1, i.e. he can change d to c, but he cannot change c to d or d to b.
2. The letter  a may not be reduced any further.

Each reduction in the value of any letter is counted as a single operation. Find the minimum number of operations required to convert a given string into a palindrome.


Function Description

Complete the theLoveLetterMystery function in the editor below.

theLoveLetterMystery has the following parameter(s):

string s: the text of the letter

Returns

int: the minimum number of operations


Input Format

The first line contains an integer q, the number of queries.
The next q  lines will each contain a string s.


Constraints

1  <=  q  <=  10

1  <=   | s |   <=  10^4
All strings are composed of lower case English letters, ascii[a-z], with no spaces.


Solution :



title-img 


                            Solution in C :

In   C++  :







#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main(){
    int t;
    cin >> t;
    while(t--){
        string s;
        cin >> s;
        int i = 0;
        int j = s.length()-1;
        int sol = 0;
        while(i<j){
            sol += abs(s[i]-s[j]);
            ++i;
            --j;
        }
        cout<<sol<<"\n";
    }
    return 0;
}








In  Java  :





import java.util.Scanner;


public class Solution {


	public static void main(String[] args) 
	{
		Scanner scan = new Scanner(System.in);
		int T = scan.nextInt();scan.nextLine();
		
		for(int i=0;i<T;i++)
		{
			String s = scan.nextLine();
			int count=0;
			for(int j=0;j<s.length()/2;j++)
				count+=Math.abs(s.charAt(j)-s.charAt(s.length()-1-j));
			System.out.println(count);
		}
	}
}








In   C :






#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int T,n,ans,i;
char s[20000];

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",&s);
        n=strlen(s);
        for(ans=i=0;i<n-1-i;i++)
            ans+=abs(s[i]-s[n-1-i]);
        printf("%d\n",ans);
    }
    return 0;
}








In   Python3  :






def case(t):
    alphabet = "abcdefghijklmnopqrstuvwxyz"
    rev = t[::-1]
    ans = 0
    for i in range(len(t) // 2):
        op1 = t[i]
        op2 = rev[i]
        pos1 = alphabet.find(op1)
        pos2 = alphabet.find(op2)
        ans += abs(pos1 - pos2)
    return ans


t = int(input())
for i in range(t):
    s = input()
    print(case(s))








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