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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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971 Discussions

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  • soundarraja_2201
     + 0 comments

    Simple C Solution

    int theLoveLetterMystery(char* s) {
    int i=0,j=strlen(s)-1,count=0;
    while(i<j){
    if(s[i]!=s[j]) count +=abs(s[i]-s[j]);
    i++;
    j--;
    }
    return count;
}

  • 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;
}

  • ariwenn15
     + 0 comments

    Java

    int changeCount = 0;
for (int i = 0; i < s.length() / 2; i++) {
    changeCount += Math.abs(s.charAt(i) - s.charAt(s.length() - i - 1));
}

return changeCount;

  • y_shchegolev1982
     + 0 comments

    return sum(abs(ord(s[i]) - ord(s[-i -1])) for i in range(int(len(s) / 2)))

  • jabarsadiqsyedi1
     + 0 comments

    JS:

    function theLoveLetterMystery(s) {
    // Write your code here
    
        let count=0;
        let c=0;
        let d=s.length-1;
        while(c<d){
             let k=s.charCodeAt(c);
             let l=s.charCodeAt(d); 
            count+=Math.abs(l-k); 
            c++;
            d--;
        }        
        return count;
}