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 adictionary
of words.boolean isUnique(string word)
Returnstrue
if either of the following conditions are met (otherwise returnsfalse
):- There is no word in
dictionary
whose abbreviation is equal toword
's abbreviation. - For any word in
dictionary
whose abbreviation is equal toword
's abbreviation, that word andword
are the same.
- There is no word in
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 toisUnique
.
Companies: Google
Related Topics:
Array, Hash Table, String, Design
Similar Questions:
// OJ: https://leetcode.com/problems/unique-word-abbreviation
// Author: github.com/lzl124631x
// Time:
// ValidWordAbbr: O(NL) where N is the length of A and L is the maximum length of A[i]
// isUnique: O(1)
// Space: O(NL)
class ValidWordAbbr {
unordered_map<string, int> cnt;
unordered_set<string> seen;
string abbr(string &s) {
return s.size() >= 3 ? s[0] + to_string(s.size() - 1) + s.back() : s;
}
public:
ValidWordAbbr(vector<string>& A) {
for (auto &s : A) {
if (seen.count(s)) continue;
cnt[abbr(s)]++;
seen.insert(s);
}
}
bool isUnique(string word) {
auto s = abbr(word);
return cnt.count(s) == 0 || (seen.count(word) && cnt[s] == 1);
}
};