Unique Morse Code Words - Practice Coding Problems

Unique Morse Code Words - Problem

International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes. 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. For example, "cab" can be written as "-.-..--...", which is the concatenation of "-.-.", ".-", and "-...".

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

Input & Output

Example 1 — Basic Case

$ Input: words = ["gin", "zen", "gig"]

› Output: 2

💡 Note: gin → '--..-..' (g='--.', i='..', n='-.'). zen → '--..-.' (z='--..', e='.', n='-.'). gig → '--..--.' (g='--.', i='..', g='--.'). gin and gig have the same morse transformation '--..-..' (wait, that's wrong). Let me recalculate: gin → '--..-.', zen → '--..-.' gig → '--..--.' So we have 2 unique transformations.

Example 2 — All Different

$ Input: words = ["a", "b", "c"]

› Output: 3

💡 Note: a → '.-', b → '-...', c → '-.-.'. All three morse codes are different, so 3 unique transformations.

Example 3 — All Same

$ Input: words = ["cab", "bac", "abc"]

› Output: 1

💡 Note: All three words contain the same letters (a, b, c), so they all produce the same morse transformation when concatenated: '-.-..-...-'

Constraints

1 ≤ words.length ≤ 100
1 ≤ words[i].length ≤ 12
words[i] consists of lowercase English letters only

Visualization

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Understanding the Visualization

Input

Array of words: ["gin", "zen", "gig"]

Process

Convert each word to morse code using the standard table

Output

Count of unique morse transformations: 2

Key Takeaway

🎯 Key Insight: Use a hash set to automatically count unique morse transformations efficiently

Asked in

G Google 15 A Amazon 12 F Facebook 8

The key insight is to convert each word to its morse code representation and count unique patterns. The optimal approach uses a hash set to automatically eliminate duplicates while adding morse transformations. Time: O(n*m), Space: O(n*k) where n is number of words, m is average word length, and k is average morse length.

Common Approaches

Approach	Time	Space	Notes
✓ Hash Set Optimization	O(n*m)	O(n*k)	Use a hash set to automatically track unique morse code transformations
Brute Force with List	O(n²)	O(n*k)	Convert each word to morse code and store in a list, then count unique transformations

Hash Set Optimization — Algorithm Steps

Create a hash set to store unique transformations
For each word, convert to morse code
Add morse code to set (duplicates ignored automatically)
Return size of set

Visualization

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Step-by-Step Walkthrough

Convert Words

Transform each word to morse code

Add to Set

Hash set automatically eliminates duplicates

Count Size

Set size equals unique transformations

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char words[][50], int n) {
    char* morse[26] = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
    
    char transformations[100][200];
    int uniqueCount = 0;
    
    for (int i = 0; i < n; i++) {
        strcpy(transformations[i], "");
        for (int j = 0; words[i][j] != '\0'; j++) {
            strcat(transformations[i], morse[words[i][j] - 'a']);
        }
        
        // Check if this transformation already exists
        int found = 0;
        for (int k = 0; k < uniqueCount; k++) {
            if (strcmp(transformations[i], transformations[k]) == 0) {
                found = 1;
                break;
            }
        }
        if (!found) {
            strcpy(transformations[uniqueCount], transformations[i]);
            uniqueCount++;
        }
    }
    
    return uniqueCount;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    char words[100][50];
    int n = 0;
    
    char* token = strtok(line, "[],\"\n ");
    while (token != NULL) {
        if (strlen(token) > 0) {
            strcpy(words[n], token);
            n++;
        }
        token = strtok(NULL, "[],\"\n ");
    }
    
    int result = solution(words, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n*m)

n words, each of average length m. Converting each word takes O(m) time

✓ Linear Growth

Space Complexity

O(n*k)

Set stores up to n unique morse transformations, each of average length k

⚡ Linearithmic Space

125.0K Views

Medium Frequency

~15 min Avg. Time

2.2K Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Hash Set Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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