LeetCode 804. Unique Morse Code Words

Description

https://leetcode.com/problems/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.

Explanation

Just implement as the problem description is.

Python Solution

class Solution:
    def uniqueMorseRepresentations(self, words: List[str]) -> int:
        results = []
        
        code_map = [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
        
        for word in words:
            word_code = ""
            
            for c in word:
                word_code += (code_map[ord(c) - ord('a')])
                        
            if word_code not in results:
                results.append(word_code)
                
        return len(results)
  • Time Complexity: O(N)
  • Space Complexity: O(N)

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