Leetcode #288: Unique Word Abbreviation
In this guide, we solve Leetcode #288 Unique Word Abbreviation in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
The abbreviation of a word is a concatenation of its first letter, the number of characters between the first and last letter, and its last letter. If a word has only two characters, then it is an abbreviation of itself.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Design, Array, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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, dictionary word "deer" and word "dear" have the same abbreviation "d2r" but are not the same.
validWordAbbr.isUnique("cart"); // return true, no words in the dictionary have the abbreviation "c2t".
validWordAbbr.isUnique("cane"); // return false, dictionary word "cake" and word "cane" have the same abbreviation "c2e" but are not the same.
validWordAbbr.isUnique("make"); // return true, no words in the dictionary have the abbreviation "m2e".
validWordAbbr.isUnique("cake"); // return true, because "cake" is already in the dictionary and no other word in the dictionary has "c2e" abbreviation.
Python Solution
class ValidWordAbbr:
def __init__(self, dictionary: List[str]):
self.d = defaultdict(set)
for s in dictionary:
self.d[self.abbr(s)].add(s)
def isUnique(self, word: str) -> bool:
s = self.abbr(word)
return s not in self.d or all(word == t for t in self.d[s])
def abbr(self, s: str) -> str:
return s if len(s) < 3 else s[0] + str(len(s) - 2) + s[-1]
# Your ValidWordAbbr object will be instantiated and called as such:
# obj = ValidWordAbbr(dictionary)
# param_1 = obj.isUnique(word)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.