Unique Word Abbreviation Python Solution

lcpy gen -n 288   # by problem number
lcpy gen -s unique_word_abbreviation   # by problem name

Input
["ValidWordAbbr", "isUnique", "isUnique", "isUnique", "isUnique", "isUnique"]
[[["deer", "door", "cake", "card"]], ["dear"], ["cart"], ["cane"], ["make"], ["cake"]]
Output
[null, false, true, false, true, true]

Explanation
ValidWordAbbr validWordAbbr = new ValidWordAbbr(["deer", "door", "cake", "card"]);
validWordAbbr.isUnique("dear"); // return false, "deer" and "dear" share "d2r" but differ.
validWordAbbr.isUnique("cart"); // return true, no dictionary word abbreviates to "c2t".
validWordAbbr.isUnique("cane"); // return false, "cake" and "cane" share "c2e" but differ.
validWordAbbr.isUnique("make"); // return true, no dictionary word abbreviates to "m2e".
validWordAbbr.isUnique("cake"); // return true, "cake" is the only word with "c2e".

from collections import defaultdict


class ValidWordAbbr:
    # Time: __init__ O(n), is_unique O(1) where n is the dictionary size
    # Space: O(n) for the abbreviation-to-words map
    def __init__(self, dictionary: list[str]) -> None:
        self.groups: dict[str, set[str]] = defaultdict(set)
        for word in dictionary:
            self.groups[self.abbr(word)].add(word)

    # Time: O(1)
    # Space: O(1)
    def is_unique(self, word: str) -> bool:
        group = self.groups.get(self.abbr(word))
        return group is None or group == {word}

    # Time: O(1)
    # Space: O(1)
    def abbr(self, word: str) -> str:
        return word if len(word) < 3 else word[0] + str(len(word) - 2) + word[-1]