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[lintcode] problem 648 – unique word abbreviation

admin 11月 13, 2020 0

An abbreviation of a word follows the form<first letter><number><last letter>. Below are some examples of word abbreviations:

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a) it                      --> it    (no abbreviation)

     1
b) d|o|g                   --> d1g

              1    1  1
     1---5----0----5--8
c) i|nternationalizatio|n  --> i18n

              1
     1---5----0
d) l|ocalizatio|n          --> l10n

Assume you have a dictionary and given a word, find whether its abbreviation is unique in the dictionary. A word’s abbreviation is unique if no other word from the dictionary has the same abbreviation.

Example

No.1

Input:
[ “deer”, “door”, “cake”, “card” ]
isUnique(“dear”)
isUnique(“cart”)

Output:
false
true

Explanation:
Dictionary’s abbreviation is [“d2r”, “d2r”, “c2e”, “c2d”].
“dear” ‘s abbreviation is “d2r” , in dictionary.
“cart” ‘s abbreviation is “c2t” , not in dictionary.

No.2

Input:
[ “deer”, “door”, “cake”, “card” ]
isUnique(“cane”)
isUnique(“make”)

Output:
false
true

Explanation:
Dictionary’s abbreviation is [“d2r”, “d2r”, “c2e”, “c2d”].
“cane” ‘s abbreviation is “c2e” , in dictionary.
“make” ‘s abbreviation is “m2e” , not in dictionary.

Code

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Map<String, Set<String>> map = new HashMap<>();

public (String[] dictionary) {
    for (String word : dictionary) {
        String abbr = getAbbreviation(word);
        map.putIfAbsent(abbr, new HashSet<>());
        map.get(abbr).add(word);
    }
}

public boolean isUnique(String word) {
    String abbr = getAbbreviation(word);
    return !map.containsKey(abbr) || (map.get(abbr).size() == 1 && map.get(abbr).contains(word));
}

private String getAbbreviation(String word) {
    return word.length() < 3 ? word : word.charAt(0) + String.valueOf(word.length() - 2) + word.charAt(word.length() - 1);
}
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