The Love-Letter Mystery

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  • + 0 comments

    Java:

    public static int theLoveLetterMystery(String s) {
      int n = s.length();
      int totalOps = 0;
    
      for (int i = 0; i < n / 2; i++) {
        totalOps += Math.abs(s.charAt(i) - s.charAt(n - 1 - i));
      }
    
      return totalOps;
    }
    
  • + 0 comments
    def theLoveLetterMystery(s):
        move=0
        for i in range(len(s)//2):
            move+=abs(ord(s[i])-ord(s[-i-1]))
        return move
    
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    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;
    }
    
  • + 0 comments

    Java solution:

         int count = 0;
         int start = 0;
         int end = s.length() - 1;
         while(true){
           if(start >= end)
             return count;
           int diff = Math.abs(s.charAt(start) - s.charAt(end));
           count += diff;
           start++;
           end--;
           }
    
  • + 0 comments

    Solution in PHP:

    function theLoveLetterMystery($s) {
        // Write your code here
        $abjad = ["a"=>1, "b"=>2, "c"=>3, "d"=>4, "e"=>5, "f"=>6, "g"=>7, "h"=>8, "i"=>9, "j"=>10, "k"=>11, "l"=>12, "m"=>13, "n"=>14, "o"=>15, "p"=>16, "q"=>17, "r"=>18, "s"=>19, "t"=>20, "u"=>21, "v"=>22, "w"=>23, "x"=>24, "y"=>25, "z"=>26];
    
        $split_s = str_split($s);
        $s_count = count($split_s);
        $separate = $s_count/2;
        $hasil = 0;
        
        if ($s_count % 2 == 0) {
            for ($i = $separate ; $i < $s_count; $i++) {
                $hasil += abs($abjad[$s[$i]] - $abjad[$s[$s_count - 1 - $i]]);
            }
        }
        else {
            for ($i = ceil($separate) ; $i < $s_count; $i++) {
                $hasil += abs($abjad[$s[$i]] - $abjad[$s[$s_count - 1 - $i]]);
            }
        }
        
        return $hasil;
    }