288. Unique Word Abbreviation 🔒

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 $O(n)$.

Python3JavaC++GoTypeScript

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)

class ValidWordAbbr {
    private Map<String, Set<String>> 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);
 */

class ValidWordAbbr {
public:
    ValidWordAbbr(vector<string>& 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<string, unordered_set<string>> 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);
 */

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);
 */

class ValidWordAbbr {
    private d: Map<string, Set<string>> = 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)
 */

288. Unique Word Abbreviation 🔒

Description

Solutions

Solution 1: Hash Table

Comments