# 734 Sentence Similarity

## 734. Sentence Similarity

## 1. Question

Given two sentences`words1, words2`(each represented as an array of strings), and a list of similar word pairs`pairs`, determine if two sentences are similar.

For example, "great acting skills" and "fine drama talent" are similar, if the similar word pairs are`pairs = [["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 sentences`words1 = ["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 like`words1 = ["great"]`can never be similar to`words2 = ["doubleplus","good"]`.

**Note:**

The length of`words1`and`words2`will not exceed`1000`.

The length of`pairs`will not exceed`2000`.

The length of each`pairs[i]`will be`2`.

The length of each`words[i]`and`pairs[i][j]`will be in the range`[1, 20]`.

## 2. Implementation

**(1) Hash Table**

```java
class Solution {
    public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) {
        if (words1.length != words2.length) {
            return false;
        }

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

        for (String[] pair : pairs) {
            synonyms.putIfAbsent(pair[0], new HashSet<>());
            synonyms.get(pair[0]).add(pair[1]);
        }

        Set<String> emptySet = new HashSet<>();

        for (int i = 0; i < words1.length; i++) {
            if (!words1[i].equals(words2[i]) && !synonyms.getOrDefault(words1[i], emptySet).contains(words2[i]) 
                && !synonyms.getOrDefault(words2[i], emptySet).contains(words1[i])) {
                return false;
            }
        }
        return true;
    }
}
```

## 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)


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