HACKER RANK Algorithm Strings Challenges - The Love Letter Mystery Solution.'C++'

Algorithm Strings Challenges - The Love Letter Mystery Solution

Problem Statement

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≤T≤10
1≤ length of string ≤104
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.

Solution:

 #include <cmath>  
 #include <cstdio>  
 #include <vector>  
 #include <iostream>  
 #include <cstring>  
 #include <algorithm>  
 using namespace std;  
 int main() {  
   /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
   int t;  
   cin>>t;  
   while(t--)  
   {  
     char inp[10000]  ;  
     cin>>inp;  
     int first=0,last=strlen(inp)-1,ans = 0,count;  
     do{  
       int ff = inp[first];  
       int ll = inp[last];  
       if(ff!=ll)  
         {  
         int count = abs (ff-ll);  
         ans = ans +count;  
       }  
       first++;  
       last--;  
     }while(first<last);  
     cout<<ans<<endl;  
   }  
   return 0;  
 } 

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