# [288. Unique Word Abbreviation](https://leetcode.com/problems/unique-word-abbreviation)
[中文文档](/solution/0200-0299/0288.Unique%20Word%20Abbreviation/README.md)
## Description
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.
For example:
dog --> d1g
because there is one letter between the first letter 'd'
and the last letter 'g'
.
internationalization --> i18n
because there are 18 letters between the first letter 'i'
and the last letter 'n'
.
it --> it
because any word with only two characters is an abbreviation of itself.
Implement the ValidWordAbbr
class:
ValidWordAbbr(String[] dictionary)
Initializes the object with a dictionary
of words.
boolean isUnique(string word)
Returns true
if either of the following conditions are met (otherwise returns false
):
- There is no word in
dictionary
whose abbreviation is equal to word
's abbreviation.
- For any word in
dictionary
whose abbreviation is equal to word
's abbreviation, that word and word
are the same.
Example 1:
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.
Constraints:
1 <= dictionary.length <= 3 * 104
1 <= dictionary[i].length <= 20
dictionary[i]
consists of lowercase English letters.
1 <= word.length <= 20
word
consists of lowercase English letters.
- At most
5000
calls will be made to isUnique
.
## Solutions
### Solution 1: Hash Table
According to the problem description, we define a function $abbr(s)$, which calculates the abbreviation of the word $s$. If the length of the word $s$ is less than $3$, then its abbreviation is itself; otherwise, its abbreviation is its first letter + (its length - 2) + its last letter.
Next, we define a hash table $d$, where the key is the abbreviation of the word, and the value is a set, the elements of which are all words abbreviated as that key. We traverse the given word dictionary, and for each word $s$ in the dictionary, we calculate its abbreviation $abbr(s)$, and add $s$ to $d[abbr(s)]$.
When judging whether the word $word$ meets the requirements of the problem, we calculate its abbreviation $abbr(word)$. If $abbr(word)$ is not in the hash table $d$, then $word$ meets the requirements of the problem; otherwise, we judge whether there is only one element in $d[abbr(word)]$. If there is only one element in $d[abbr(word)]$ and that element is $word$, then $word$ meets the requirements of the problem.
In terms of time complexity, the time complexity of initializing the hash table is $O(n)$, where $n$ is the length of the word dictionary; the time complexity of judging whether a word meets the requirements of the problem is $O(1)$. In terms of space complexity, the space complexity of the hash table is $O(n)$.
```python
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)
```
```java
class ValidWordAbbr {
private Map> d = new HashMap<>();
public ValidWordAbbr(String[] dictionary) {
for (var s : dictionary) {
d.computeIfAbsent(abbr(s), k -> new HashSet<>()).add(s);
}
}
public boolean isUnique(String word) {
var ws = d.get(abbr(word));
return ws == null || (ws.size() == 1 && ws.contains(word));
}
private String abbr(String s) {
int n = s.length();
return n < 3 ? s : s.substring(0, 1) + (n - 2) + s.substring(n - 1);
}
}
/**
* Your ValidWordAbbr object will be instantiated and called as such:
* ValidWordAbbr obj = new ValidWordAbbr(dictionary);
* boolean param_1 = obj.isUnique(word);
*/
```
```cpp
class ValidWordAbbr {
public:
ValidWordAbbr(vector& dictionary) {
for (auto& s : dictionary) {
d[abbr(s)].insert(s);
}
}
bool isUnique(string word) {
string s = abbr(word);
return !d.count(s) || (d[s].size() == 1 && d[s].count(word));
}
private:
unordered_map> d;
string abbr(string& s) {
int n = s.size();
return n < 3 ? s : s.substr(0, 1) + to_string(n - 2) + s.substr(n - 1, 1);
}
};
/**
* Your ValidWordAbbr object will be instantiated and called as such:
* ValidWordAbbr* obj = new ValidWordAbbr(dictionary);
* bool param_1 = obj->isUnique(word);
*/
```
```go
type ValidWordAbbr struct {
d map[string]map[string]bool
}
func Constructor(dictionary []string) ValidWordAbbr {
d := make(map[string]map[string]bool)
for _, s := range dictionary {
abbr := abbr(s)
if _, ok := d[abbr]; !ok {
d[abbr] = make(map[string]bool)
}
d[abbr][s] = true
}
return ValidWordAbbr{d}
}
func (this *ValidWordAbbr) IsUnique(word string) bool {
ws := this.d[abbr(word)]
return ws == nil || (len(ws) == 1 && ws[word])
}
func abbr(s string) string {
n := len(s)
if n < 3 {
return s
}
return fmt.Sprintf("%c%d%c", s[0], n-2, s[n-1])
}
/**
* Your ValidWordAbbr object will be instantiated and called as such:
* obj := Constructor(dictionary);
* param_1 := obj.IsUnique(word);
*/
```
```ts
class ValidWordAbbr {
private d: Map> = new Map();
constructor(dictionary: string[]) {
for (const s of dictionary) {
const abbr = this.abbr(s);
if (!this.d.has(abbr)) {
this.d.set(abbr, new Set());
}
this.d.get(abbr)!.add(s);
}
}
isUnique(word: string): boolean {
const ws = this.d.get(this.abbr(word));
return ws === undefined || (ws.size === 1 && ws.has(word));
}
abbr(s: string): string {
const n = s.length;
return n < 3 ? s : s[0] + (n - 2) + s[n - 1];
}
}
/**
* Your ValidWordAbbr object will be instantiated and called as such:
* var obj = new ValidWordAbbr(dictionary)
* var param_1 = obj.isUnique(word)
*/
```