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  1. Prepare
  2. Algorithms
  3. Strings
  4. The Love-Letter Mystery
  5. Discussions

The Love-Letter Mystery

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986 Discussions

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  • smoralescarrie
     + 0 comments

    This problem reminds me of how we sometimes make small changes for love—just like in frases de amor para mi esposo, where every word matters to make someone feel special. In this challenge, each letter change counts too, turning the message into something more beautiful... like a romantic palindrome!

  • geovanni_luna
     + 0 comments

    Here is a python solution in O(n) time and O(1) space:

    def theLoveLetterMystery(s):
    letters = "abcdefghijklmnopqrstuvwxyz"
    
    i = 0
    j = len(s) - 1
    
    change_count = 0
    
    while i < j:
        if s[i] == s[j]:
            i += 1
            j -= 1
        else:
            idx_i = letters.index(s[i])
            idx_j = letters.index(s[j])
            temp_count = abs(idx_j - idx_i)
            
            change_count += temp_count
            i += 1
            j -= 1
        
    return change_count

  • cass6896
     + 0 comments

    C++

    int theLoveLetterMystery(string s)
{
    int count{0};
    for (int i = s.length() - 1; i >= s.length() / 2; i--)
        count += abs(s[i] - s[s.length() - 1 - i]);
        
    return count;
}

  • ahmet_alp77
     + 0 comments

    Java:

    public static int theLoveLetterMystery(String s) {
    int left = 0;
    int right = s.length() - 1;
    int retVal = 0;
    while (left < right) {
        if (s.charAt(left) != s.charAt(right)) {
            retVal += Math.abs((int) s.charAt(right) - (int) s.charAt(left));
        }
        ++left;
        --right;
    }
    return retVal;
}

  • alban_tyrex
     + 0 comments

    Here is my c++ solution to the problem, explanation here : https://youtu.be/gIUqS2xO6lg

    int theLoveLetterMystery(string s) {
    int ans = 0 , start = 0, end = s.size() - 1;
    while(start < end){
        ans += abs(s[start] - s[end]);
        start++; end--;
    }
    return ans;
}