Hackerrank - The Love-Letter Mystery Solution

Hackerrank - The Love-Letter Mystery Solution

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: cdecddcdc.

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 .


| 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


Sample Output



  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.

Solution in Python

def theLoveLetterMystery(s):
    total = 0
    for i,j in enumerate(range(len(s)//2),1):
        a = s[ j]
        b = s[-i]
        total+= abs(ord(a)-ord(b))
    return tot

for _ in range(int(input())):

Additional Information

Variable a - Left part of string
Example aabb in aabbcbbaa or aabb in aabbbbaa
Variable b - Right part of string

Using deque

from collections import deque
def theLoveLetterMystery(s):
    s = deque((ord(i) for i in s))
    return sum(abs(s.pop()-s.popleft()) for i in range(len(s)//2))

for _ in range(int(input())):

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