Unique Morse Code Words

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='--.'). We have gin and zen with the same morse transformation '--..-..' and gig with a different one. 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: cab → '-.-..--...', bac → '-...-.-..-', abc → '.--...-.-.' All three words produce different morse transformations when the order matters, so we have 3 unique transformations, not 1.

Constraints

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

Visualization

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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

✓ Brute Force with List

⏱️ Time: O(n²) Space: O(n*k)

For each word, convert it to morse code by concatenating morse codes of individual letters. Store all transformations in a list and manually count unique ones.

Hash Set Optimization

⏱️ Time: O(n*m) Space: O(n*k)

Convert each word to its morse code representation and add it to a hash set. The set automatically handles duplicates, so the final size gives us the count of unique transformations.

Brute Force with List — Algorithm Steps

Convert each word to its morse code transformation
Store all transformations in a list
Count unique transformations by comparing each with others

Visualization

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Step-by-Step Walkthrough

Convert to Morse

Transform each word to morse code

Store All

Keep all transformations in a list

Count Unique

Compare each transformation with previous ones

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];
    
    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']);
        }
    }
    
    // Count unique transformations manually
    int uniqueCount = 0;
    for (int i = 0; i < n; i++) {
        int isUnique = 1;
        for (int j = 0; j < i; j++) {
            if (strcmp(transformations[i], transformations[j]) == 0) {
                isUnique = 0;
                break;
            }
        }
        if (isUnique) {
            uniqueCount++;
        }
    }
    
    return uniqueCount;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

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(n²)

O(n*m) for conversions where n is words count and m is average word length, plus O(n²) for counting unique transformations

✓ Linear Growth

Space Complexity

O(n*k)

List stores n transformations, each of average length k

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force with List — Algorithm Steps

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Code -

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