HackerRank The Love-Letter Mystery Solution

Yashwant Parihar,

In this post, we will solve HackerRank The Love-Letter Mystery Problem 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 1. 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.
    Example
    8 = cde
    The following two operations are performed: cde → cdd → cdc. Return 2.

Function Description

Complete the theLoveLetterMystery function in the editor below.

theLoveLetterMystery has the following parameter(s):

  • string s: the text of the letter

Returns

  • int: the minimum number of operations

Input Format

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

Sample Input

STDIN   Function
-----   --------
4       q = 4
abc     query 1 = 'abc'
abcba
abcd
cba

Sample Output

2
0
4
2

Explanation

  1. For the first query, abc → abb → aba.
  2. For the second query, abcba is already a palindromic string.
  3. For the third query, abcd→ abcc→ abcb → abca → abba.
  4. For the fourth query, cba → bba → aba.
HackerRank The Love-Letter Mystery Problem Solution
HackerRank The Love-Letter Mystery Problem Solution

The Love-Letter Mystery C Solution

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

#define MAX_STR_LEN 10000

int main(void) {
    int testCases, i, j, lenstr, numOps;
    char word[MAX_STR_LEN];
    scanf("%d", &testCases);
    if (testCases > 10 || testCases < 1){
        fprintf(stderr,"Usage: T out of bounds\n");
        return 1;
    }

    for(i = 0; i < testCases; ++i) {
        numOps = 0;
        scanf("%s", word);
        lenstr = (int) strlen(word);
        for(j = 0; j < lenstr/2; ++j) {
            numOps += abs(tolower(word[j]) - tolower(word[lenstr - j - 1]));
        }
        printf("%d\n",numOps);
    }

    return 0;
}

The Love-Letter Mystery C++ Solution

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int length;
    cin>>length;
    for(int i = 0; i < length; ++i){
        int red = 0;
        string s;
        cin>>s;
        for(int j = 0; j < s.length()/2; ++j){
            red += abs((char)s[j]-(char)s[s.length()-j-1]);
        }
        cout<<red<<endl;
    }
    return 0;
}

The Love-Letter Mystery C Sharp Solution

using System;
using System.Linq;
using System.Collections.Generic;
using System.IO;
class Solution 
{
    static void Main(String[] args) 
    {    
        int numCases = Convert.ToInt32(Console.ReadLine());
        
        for(int i = 0; i < numCases; i++)
        {
            string source = Console.ReadLine().Trim();
            int halfLength = Convert.ToInt32(Math.Floor((double)source.Length / 2.0));
            int count = 0;
            
            char[] firstHalf = source.Substring(0, halfLength).ToCharArray();
            char[] secondHalf = source.Substring(source.Length - halfLength, halfLength).ToCharArray();
            
            int secondHalfMaxIndex = secondHalf.Length-1;
            
            for(int x = 0; x < halfLength; x++)
            {
                if(firstHalf[x] == secondHalf[secondHalfMaxIndex - x]) { continue; } // do nothing
                else if (firstHalf[x] > secondHalf[secondHalfMaxIndex - x]) // decrement letter in first half
                {
                    while(firstHalf[x] > secondHalf[secondHalfMaxIndex - x])
                    {
                        firstHalf[x] = (char)( (int)firstHalf[x] - 1 );
                        count++;
                    } 
                }
                else if (firstHalf[x] < secondHalf[secondHalfMaxIndex - x]) // decrement letter in second half
                {
                    while(firstHalf[x] < secondHalf[secondHalfMaxIndex - x])
                    {
                        secondHalf[secondHalfMaxIndex - x] = (char)( (int)secondHalf[secondHalfMaxIndex - x] - 1 );
                        count++;
                    } 
                }
            }
            
            Console.WriteLine(count);
        }
    }
    
}

The Love-Letter Mystery Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'theLoveLetterMystery' function below.
     *
     * The function is expected to return an INTEGER.
     * The function accepts STRING s as parameter.
     */

    public static int theLoveLetterMystery(String s) {
    // Write your code here
        var charArr = s.toCharArray();
        var totalCount = 0;
        for (var i=0; i<s.length()/2; i++) {
            totalCount += Math.abs(charArr[i] - charArr[s.length()-1-i]);
        }
        return totalCount;

    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int q = Integer.parseInt(bufferedReader.readLine().trim());

        IntStream.range(0, q).forEach(qItr -> {
            try {
                String s = bufferedReader.readLine();

                int result = Result.theLoveLetterMystery(s);

                bufferedWriter.write(String.valueOf(result));
                bufferedWriter.newLine();
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        });

        bufferedReader.close();
        bufferedWriter.close();
    }
}

The Love-Letter Mystery JavaScript Solution

'use strict';

var alphabet = "abcdefghijklmnopqrstuvwxyz";
function diff(a, b) {
    a = alphabet.indexOf(a);
    b = alphabet.indexOf(b);
    return Math.abs(a - b);
}

function calc(word) {    
      
    var operations = 0;
    var half = Math.floor(word.length / 2);
    var l = word.length-1;
    
    if(word.length === 1) return 0;
    
    for(var i = 0; i < half; i++) {
        var a = word.charAt(i);
        var b = word.charAt(l-i);
        operations += diff(a, b);
    }
    
    return operations;
}

function processData(input) {
    
    var lines = input.split('\n');
    var ans = "";
    for(var i = 1; i < lines.length; i++) {
        ans += calc(lines[i]) + "\n";
    }
    process.stdout.write(ans);
}

process.stdin.resume();
process.stdin.setEncoding("ascii");
var _input = "";
process.stdin.on("data", function (input) { _input += input; });
process.stdin.on("end", function () { processData(_input); });

The Love-Letter Mystery Python Solution

import sys
from operator import abs

def main():
    t = sys.stdin.readline().strip()
    for i in range(int(t)):
        testcase = sys.stdin.readline().strip()
        cnt = 0
        for j in range(int(len(testcase) / 2)):
            cnt = cnt + abs(ord(testcase[j]) - ord(testcase[len(testcase) - 1 - j]))
        print(cnt)
    
if __name__ == "__main__":
    main()

Other Solutions

c C# C++ HackerRank Solutions java javascript python

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