734 Sentence Similarity

734. Sentence Similarity

1. Question

Given two sentenceswords1, words2(each represented as an array of strings), and a list of similar word pairspairs, determine if two sentences are similar.
For example, "great acting skills" and "fine drama talent" are similar, if the similar word pairs arepairs = [["great", "fine"], ["acting","drama"], ["skills","talent"]].
Note that the similarity relation is not transitive. For example, if "great" and "fine" are similar, and "fine" and "good" are similar, "great" and "good" are not necessarily similar.
However, similarity is symmetric. For example, "great" and "fine" being similar is the same as "fine" and "great" being similar.
Also, a word is always similar with itself. For example, the sentenceswords1 = ["great"], words2 = ["great"], pairs = []are similar, even though there are no specified similar word pairs.
Finally, sentences can only be similar if they have the same number of words. So a sentence likewords1 = ["great"]can never be similar towords2 = ["doubleplus","good"].
Note:
The length ofwords1andwords2will not exceed1000.
The length ofpairswill not exceed2000.
The length of eachpairs[i]will be2.
The length of eachwords[i]andpairs[i][j]will be in the range[1, 20].

2. Implementation

(1) Hash Table
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class Solution {
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public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) {
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if (words1.length != words2.length) {
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return false;
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}
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Map<String, Set<String>> synonyms = new HashMap<>();
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for (String[] pair : pairs) {
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synonyms.putIfAbsent(pair[0], new HashSet<>());
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synonyms.get(pair[0]).add(pair[1]);
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}
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Set<String> emptySet = new HashSet<>();
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for (int i = 0; i < words1.length; i++) {
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if (!words1[i].equals(words2[i]) && !synonyms.getOrDefault(words1[i], emptySet).contains(words2[i])
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&& !synonyms.getOrDefault(words2[i], emptySet).contains(words1[i])) {
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return false;
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}
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}
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return true;
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}
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}
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3. Time & Space Complexity

Hash Table: 时间复杂度O(n + k), n为words的长度,k是pairs的长度。空间复杂度是O(k)