The Love-Letter Mystery : Challenge | Strings | Algorithms | HackerRank

The Love-Letter Mystery : Challenge | Strings | Algorithms | HackerRank

James found a love letter his friend Harry has written for 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 reduce the value of a letter, e.g. he can change d to c, but he cannot change c to d.
2. In order to form a palindrome, if he has to repeatedly reduce the value of a letter, he can do it until the letter becomes a. Once a letter has been changed to a, it can no longer be changed.

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.

Input Format

The first line contains an integer $T$, i.e., the number of test cases.
The next $T$ lines will contain a string each. The strings do not contain any spaces.
Constraints
$1 \le T \le 10$
$1 \le$ length of string $\le 10^4$
All characters are lower case English letters.

Output Format

A single line containing the number of minimum operations corresponding to each test case.

Sample Input

4  abc  abcba  abcd  cba  

Sample Output

2  0  4  2  

Explanation

1. For the first test case, abc -> abb -> aba.
2. For the second test case, abcba is already a palindromic string.
3. For the third test case, abcd -> abcc -> abcb -> abca = abca -> abba.
4. For the fourth test case, cba -> bba -> aba.

https://sisijava.wordpress.com/2014/07/14/hackerrank-the-love-letter-mystery/

            int m = 0;

            int n = s_letters.length-1;

            while(m<n){

                if(s_letters[m]!=s_letters[n]){

                    int tmp = Math.abs(s_letters[m]-s_letters[n]);

                    count = count+tmp;

 

                }

                m++;

                n--;

            }

https://github.com/havelessbemore/hackerrank/blob/master/algorithms/warmup/the-love-letter-mystery.java
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