Problem Description

This problem asks you to implement a class that checks if a word's abbreviation is unique within a dictionary.

Abbreviation Rules:

• For words with 3 or more characters: the abbreviation is the first letter + the count of middle characters + the last letter
  • Example: "dog" becomes "d1g" (1 character between 'd' and 'g')
  • Example: "internationalization" becomes "i18n" (18 characters between 'i' and 'n')
• For words with exactly 2 characters: the word is its own abbreviation
  • Example: "it" remains "it"

Class Implementation:

You need to implement the ValidWordAbbr class with two methods:

1. __init__(dictionary): Initialize the object with an array of words
2. isUnique(word): Returns true if the word is unique, false otherwise

Uniqueness Definition:

A word is considered unique if ONE of these conditions is true:

• No word in the dictionary has the same abbreviation as the given word
• All words in the dictionary that share the same abbreviation are exactly the same as the given word

For example, if the dictionary contains ["deer", "door", "cake", "card"]:

• isUnique("dear") returns false because "deer" from the dictionary has the same abbreviation "d2r"
• isUnique("cart") returns true because although "card" has abbreviation "c2d", "cart" has abbreviation "c2t" which doesn't match
• isUnique("cane") returns false because both "cake" and "cane" have abbreviation "c2e" but they are different words
• isUnique("make") returns true because no word in dictionary has abbreviation "m2e"
• isUnique("cake") returns true because although "cake" in dictionary has abbreviation "c2e", it's the same word

Intuition

The key insight is that we need to quickly check if a word's abbreviation conflicts with words in our dictionary. Since we'll be making multiple queries, we should preprocess the dictionary to make lookups efficient.

Think about what we're really checking: given a word, we want to know if its abbreviation is "unique" according to the specific rules. The word is unique if either:

1. No dictionary word shares its abbreviation, OR
2. All dictionary words that share its abbreviation are identical to the query word itself

This suggests we should group dictionary words by their abbreviations. A hash table is perfect for this - we can use the abbreviation as the key and store all words with that abbreviation as the value.

But what data structure should we use for the value? Since we need to:

• Check if all words with a given abbreviation are the same
• Avoid duplicate entries if the same word appears multiple times in the dictionary

A set is ideal. It automatically handles duplicates and makes it easy to check if all elements are identical to our query word.

During initialization, we compute each word's abbreviation and add it to our hash table. For the abbreviation function, we handle the special case where words with less than 3 characters are their own abbreviation.

When checking if a word is unique, we:

1. Compute its abbreviation
2. If that abbreviation isn't in our hash table, the word is unique
3. If it is in our hash table, we check if all words in that set are identical to our query word

This approach gives us O(1) average time complexity for lookups after an O(n) preprocessing step, where n is the size of the dictionary.

Solution Approach

We implement the solution using a hash table with sets as values to efficiently group words by their abbreviations.

Helper Function - abbr(s): First, we create a helper function to compute abbreviations:

• If the word length is less than 3, return the word itself
• Otherwise, return s[0] + str(len(s) - 2) + s[-1]

Initialization - __init__(dictionary):

1. Create a defaultdict(set) called d to store our mapping
2. Iterate through each word s in the dictionary
3. Calculate the abbreviation using self.abbr(s)
4. Add the word to the set at d[abbreviation]

Using a defaultdict(set) ensures:

• No key errors when accessing non-existent abbreviations
• Automatic deduplication if the same word appears multiple times
• Easy grouping of all words with the same abbreviation

Uniqueness Check - isUnique(word):

1. Calculate the abbreviation of the query word: s = self.abbr(word)
2. Check if this abbreviation exists in our hash table
3. Return true if either:
   • The abbreviation s is not in d (no conflicts), OR
   • All words in d[s] are identical to word (checked using all(word == t for t in self.d[s]))

The expression all(word == t for t in self.d[s]) efficiently verifies that every word in the set with abbreviation s is exactly the same as our query word. This handles the case where the word itself is in the dictionary.

Time Complexity:

• Initialization: O(n) where n is the dictionary size
• isUnique: O(m) where m is the number of words with the same abbreviation (usually small)

Space Complexity: O(n) to store the dictionary words in the hash table

Example Walkthrough

Let's walk through a small example with dictionary ["deer", "door", "cake", "card"].

Step 1: Initialization

We create our hash table d and process each word:

1. "deer" → abbreviation: "d2r" (d + 2 middle chars + r)

   • d["d2r"] = {"deer"}

2. "door" → abbreviation: "d2r" (d + 2 middle chars + r)

   • d["d2r"] = {"deer", "door"}

3. "cake" → abbreviation: "c2e" (c + 2 middle chars + e)

   • d["c2e"] = {"cake"}

4. "card" → abbreviation: "c2d" (c + 2 middle chars + d)

   • d["c2d"] = {"card"}

Final hash table:

d = {
    "d2r": {"deer", "door"},
    "c2e": {"cake"},
    "c2d": {"card"}
}

Step 2: Query Examples

Query 1: isUnique("dear")

• Abbreviation of "dear" = "d2r"
• Check: Is "d2r" in d? Yes, it maps to {"deer", "door"}
• Check: Are all words in {"deer", "door"} equal to "dear"? No
• Return: false

Query 2: isUnique("cart")

• Abbreviation of "cart" = "c2t"
• Check: Is "c2t" in d? No
• Return: true (no conflicts)

Query 3: isUnique("cake")

• Abbreviation of "cake" = "c2e"
• Check: Is "c2e" in d? Yes, it maps to {"cake"}
• Check: Are all words in {"cake"} equal to "cake"? Yes
• Return: true (all words with this abbreviation are identical to query)

Query 4: isUnique("door")

• Abbreviation of "door" = "d2r"
• Check: Is "d2r" in d? Yes, it maps to {"deer", "door"}
• Check: Are all words in {"deer", "door"} equal to "door"? No ("deer" ≠ "door")
• Return: false

This walkthrough demonstrates how the hash table efficiently groups words by abbreviation and how the uniqueness check correctly handles both cases: when no conflicts exist and when all conflicting words are identical to the query.

Solution Implementation

Time and Space Complexity

Time Complexity:

• Initialization (__init__): O(n) where n is the total number of words in the dictionary. We iterate through each word once, and for each word, we compute its abbreviation in O(1) time (accessing first and last characters and calculating length) and add it to a set in O(1) average time.

• isUnique method: O(m) where m is the number of words in the set self.d[s]. In the worst case, we need to check all words in the set to verify if they all equal the input word. However, in practice, m is typically small, making this close to O(1) for most cases.

• abbr method: O(1) as it only accesses the first and last characters of the string and performs basic arithmetic operations.

Space Complexity:

• O(n) where n is the total number of unique words in the dictionary. The hash table stores at most n words distributed across different abbreviation keys. Each abbreviation key maps to a set of words, and the total number of words stored across all sets is exactly n.

Common Pitfalls

1. Not Handling Duplicate Words in Dictionary

The Pitfall: Many implementations fail to account for duplicate words in the input dictionary. If the dictionary contains ["dog", "dog"], storing them in a list would incorrectly count "dog" twice when checking uniqueness, leading to wrong results.

Why It Matters: When checking isUnique("dog") with dictionary ["dog", "dog"], a naive list-based approach might conclude it's not unique because it finds multiple entries, even though they're all the same word.

The Solution: Use a set instead of a list to store words for each abbreviation. This automatically handles duplicates:

# Correct: Using set
self.abbreviation_map = defaultdict(set)
self.abbreviation_map[abbreviation].add(word)  # Duplicates automatically handled

# Incorrect: Using list
self.abbreviation_map = defaultdict(list)
self.abbreviation_map[abbreviation].append(word)  # Would store duplicates

2. Incorrect Abbreviation for Short Words

The Pitfall: Forgetting that words with length less than 3 should not be abbreviated. A common mistake is trying to create abbreviations like "a1" for "ab" or accessing out-of-bounds indices.

Why It Matters: The problem specifically states that 2-character words remain unchanged. Attempting to abbreviate "it" as "i0t" would be incorrect.

The Solution: Always check the word length before creating an abbreviation:

def _get_abbreviation(self, word: str) -> str:
    if len(word) < 3:  # Critical check
        return word
    return f"{word[0]}{len(word) - 2}{word[-1]}"

3. Misunderstanding the Uniqueness Condition

The Pitfall: A common misinterpretation is checking if the word exists in the dictionary rather than checking if its abbreviation conflicts with different words. Some incorrectly implement:

# WRONG: This checks if the word is in dictionary
return word not in self.all_words

Why It Matters: The uniqueness is about abbreviation conflicts, not word existence. If dictionary has ["deer"] and we check isUnique("door"), it should return false because both abbreviate to "d2r", even though "door" isn't in the dictionary.

The Solution: Check if all words with the same abbreviation are identical to the query word:

# Correct: Check if all words with same abbreviation are the same as input
return all(existing_word == word for existing_word in self.abbreviation_map[abbreviation])

4. Not Handling Empty Abbreviation Groups

The Pitfall: When an abbreviation doesn't exist in the map, accessing it without proper handling could cause errors or incorrect logic.

Why It Matters: If we check isUnique("make") and no word has abbreviation "m2e", we need to return true without errors.

The Solution: Either use defaultdict(set) or explicitly check for key existence:

# Solution 1: Using defaultdict (preferred)
self.abbreviation_map = defaultdict(set)

# Solution 2: Explicit checking
return (abbreviation not in self.abbreviation_map or 
        all(word == w for w in self.abbreviation_map[abbreviation]))