Leetcode #804: Unique Morse Code Words
In this guide, we solve Leetcode #804 Unique Morse Code Words in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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: [".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."] Given an array of strings words where each word can be written as a concatenation of the Morse code of each letter.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Array, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
[".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."]
Python Solution
class Solution:
def uniqueMorseRepresentations(self, words: List[str]) -> int:
codes = [
".-",
"-...",
"-.-.",
"-..",
".",
"..-.",
"--.",
"....",
"..",
".---",
"-.-",
".-..",
"--",
"-.",
"---",
".--.",
"--.-",
".-.",
"...",
"-",
"..-",
"...-",
".--",
"-..-",
"-.--",
"--..",
]
s = {''.join([codes[ord(c) - ord('a')] for c in word]) for word in words}
return len(s)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.