The Love Letter Mystery Problem

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 intopalindromes.

To do this, he follows 2 rules:

(a) He can reduce the value of a letter, e.g. he can change 'd' to 'c', but he cannot change 'c' to 'd'. 
(b) 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.

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

Constraints 
1 ≤ T ≤ 10
1 ≤ length of string ≤ 104 
All characters are lower case English letters.

Sample Input #00
3
abc
abcba
abcd


Sample Output #00
2
0
4


Explanation

For the first test case, ab*c* -> ab*b* -> ab*a*. 
For the second test case, abcba is a palindromic string. 
For the third test case, abc*d* -> abc*c* -> abc*b* -> abc*a* = ab*c*a -> ab*b*a.

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

int main() {
    int T,i,j,len,count=0;
    char **s;
    scanf("%d",&T);
    s=(char**)malloc(sizeof(char*)*T);
    
    for(i=0;i<T;i++)
    { 
    s[i]=(char*)malloc(sizeof(char)*10000);   
    scanf("%s",s[i]);
    }
    for(i=0;i<T;i++)
     {   count=0;
        len=strlen(s[i]);
        for(j=0;j<len/2;j++)
        {
          if(s[i][j]>s[i][len-j-1])
          {
          while(s[i][j]!=s[i][len-j-1])
          {
           s[i][j]=(char)((int)s[i][j]-1);   
           count++;    
          }    
          }
          else
          if(s[i][j]<s[i][len-j-1])
          {
            while(s[i][j]!=s[i][len-j-1])
          {
           s[i][len-j-1]=(char)((int)s[i][len-j-1]-1);   
           count++;    
          }   
              
          }        
        }
      printf("%d\n",count);
           
    }   
      
    return 0;
}