0868EasyArrayHash TableString

Distinct Flag Strings for the Signal Words

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A signal book gives each lowercase letter a fixed flag pattern. Letter 'a' uses the pattern ".-", and the twenty-six patterns in letter order are

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

A word is signalled by writing the patterns of its letters one after another with nothing between them, which is called its flag string.

Return how many different flag strings the words of words produce.

Examples

Example 1

Input: words = ["ee", "i"]
Output: 1

Letter "e" is a single dot and letter "i" is two dots, so both words signal ".." and only one flag string comes out.

Example 2

Input: words = ["a", "b"]
Output: 2

The two letters have different patterns, so the two words signal different flag strings.

Example 3

Input: words = ["ab", "ab", "ba"]
Output: 2

The two copies of the first word signal the same flag string, and reversing the letters gives a different one, so two come out.

Constraints

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

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def distinct_flag_strings(words: list[str]) -> int:

Java

public int distinctFlagStrings(String[] words)

The intended approach

Time O(total letters) across all words: each letter contributes a fixed-length pattern.Space O(total pattern length) for the set of flag strings.

Look each letter's pattern up by its offset from the first letter of the alphabet, join a word's patterns into one string, and drop the results into a set. The size of the set is the count of different flag strings, since a set keeps one copy of each.

Pythonentry point: distinct_flag_strings

def distinct_flag_strings(words):
    book = [".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---",
            "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-",
            "..-", "...-", ".--", "-..-", "-.--", "--.."]
    seen = set()
    for word in words:
        seen.add("".join(book[ord(ch) - 97] for ch in word))
    return len(seen)

The obvious version

Build each word's flag string by walking its letters and searching the book for the matching letter, then collect the strings in a list and count how many are not equal to an earlier one.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_distinct_flag_strings(words):
    alpha = "abcdefghijklmnopqrstuvwxyz"
    book = [".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---",
            "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-",
            "..-", "...-", ".--", "-..-", "-.--", "--.."]
    built = []
    for word in words:
        text = ""
        for ch in word:
            for i in range(26):
                if alpha[i] == ch:
                    text += book[i]
                    break
        built.append(text)
    count = 0
    for i in range(len(built)):
        fresh = True
        for j in range(i):
            if built[j] == built[i]:
                fresh = False
                break
        if fresh:
            count += 1
    return count

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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