[LeetCode&Python] Problem 804. Unique Morse Code Words

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:

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

Now, given a list of words, each word can be written as a concatenation of the Morse code of each letter. For example, "cab" 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:
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 "--...--.".

Note:

• The length of words will be at most 100.
• Each words[i] will have length in range [1, 12].
• words[i] will only consist of lowercase letters.

---

 

 

Though every alphabet represents different Morse Code, some of the combinations will be the same. To solve this problem, the first step is that we need to change all the strings to their Morse Codes. The second step is to store the code in a list. If the code is already in the list, we do not store it. After the second step, we just need to count the lenght of the code list. Then the lenght is the answer.

 

class Solution:
    def uniqueMorseRepresentations(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        sta=[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]

        w=[]

        for s in words:
            a=''
            for c in s:
                a=a+sta[ord(c)-ord('a')]
            if not a in w:
                w.append(a)
        return len(w)