Question:

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: "a" maps to ".-", "b" maps to "-...", "c" maps to "-.-.", and so on.

For convenience, the full table for the 26 letters of the English alphabet is given below:

  1. [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

Now, given a list of words, each word can be written as a concatenation of the Morse code of each letter. For example, “cba” can be written as “-.-..—…”, (which is the concatenation “-.-.” + “-…” + “.-“). We’ll call such a concatenation, the transformation of a word.

Return the number of different transformations among all words we have.

Example:

Example:
Input: words = ["gin", "zen", "gig", "msg"]
Output: 2
Explanation: 
The transformation of each word is:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."

There are 2 different transformations, "--...-." and "--...--.".

Solution:

/**
 * @param {string[]} words
 * @return {number}
 */
var uniqueMorseRepresentations = function(words) {
  const mosi = {
    a:".-",
    b:"-...",
    c:"-.-.",
    d:"-..",
    e:".",
    f:"..-.",
    g:"--.",
    h:"....",
    i:"..",
    j:".---",
    k:"-.-",
    l:".-..",
    m:"--",
    n:"-.",
    o:"---",
    p:".--.",
    q:"--.-",
    r:".-.",
    s:"...",
    t:"-",
    u:"..-",
    v:"...-",
    w:".--",
    x:"-..-",
    y:"-.--",
    z:"--.."
  };
  let num = 0 ;
  let map = {};
  let temp;

  for (let i = 0; i < words.length; i++) {
    temp = words[i].split('').map(j=>mosi[j]).join('');

    if (!map[temp]) {
      map[temp]=1;
      num++;
    }

  }

  return num;
};

Runtime: 60 ms, faster than 66.61% of JavaScript online submissions for Unique Morse Code Words.