The Love-Letter Mystery HackerRank Solution

The Love-Letter Mystery HackerRank Solution
Source : https://www.hackerrank.com/challenges/the-love-letter-mystery

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 , 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.

For example, given the string , the following two operations are performed: cde → cdd → cdc.

Function Description

Complete the theLoveLetterMystery function in the editor below. It should return the integer representing the minimum number of operations needed to make the string a palindrome.

theLoveLetterMystery has the following parameter(s):

• s: a string

Input Format

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

Constraints

| s |
All strings are composed of lower case English letters, *ascii[a-z], with no spaces.

Output Format

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

Sample Input

4
abc
abcba
abcd
cba

Sample Output

2042

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 → abba.
4. For the fourth test case, cba → bba → aba.

Source : https://www.hackerrank.com/challenges/the-love-letter-mystery

Solution

	// Karthikalapati.blogspot.com
	import java.util.Scanner;
	// We compare ASCII values for pairs of characters. The 1st pair we consider is the 2
	// characters at opposite ends of the String. We then move inwards until we consider
	// all palindromic pairs.
	public class Solution {
	public static void main(String[] args) {
	Scanner scan = new Scanner(System.in);
	int T = scan.nextInt();
	while (T-- > 0) {
	String str = scan.next();
	System.out.println(minimumOperations(str));
	}
	scan.close();
	}
	private static int minimumOperations(String str) {
	int count = 0;
	for (int i = 0; i < str.length() / 2; i++) {
	count += Math.abs(str.charAt(i) - str.charAt(str.length() - 1 - i));
	}
	return count;
	}
	}