# 0737. Sentence Similarity II

https://leetcode.com/problems/sentence-similarity-ii

## Description

We can represent a sentence as an array of words, for example, the sentence `"I am happy with leetcode"` can be represented as `arr = ["I","am",happy","with","leetcode"]`.

Given two sentences `sentence1` and `sentence2` each represented as a string array and given an array of string pairs `similarPairs` where `similarPairs[i] = [xi, yi]` indicates that the two words `xi` and `yi` are similar.

Return `true` _if `sentence1` and `sentence2` are similar, or_ `false` _if they are not similar_.

Two sentences are similar if:

* They have **the same length** (i.e., the same number of words)
* `sentence1[i]` and `sentence2[i]` are similar.

Notice that a word is always similar to itself, also notice that the similarity relation is transitive. For example, if the words `a` and `b` are similar, and the words `b` and `c` are similar, then `a` and `c` are **similar**.

**Example 1:**

```
**Input:** sentence1 = ["great","acting","skills"], sentence2 = ["fine","drama","talent"], similarPairs = [["great","good"],["fine","good"],["drama","acting"],["skills","talent"]]
**Output:** true
**Explanation:** The two sentences have the same length and each word i of sentence1 is also similar to the corresponding word in sentence2.
```

**Example 2:**

```
**Input:** sentence1 = ["I","love","leetcode"], sentence2 = ["I","love","onepiece"], similarPairs = [["manga","onepiece"],["platform","anime"],["leetcode","platform"],["anime","manga"]]
**Output:** true
**Explanation:** "leetcode" --> "platform" --> "anime" --> "manga" --> "onepiece".
Since "leetcode is similar to "onepiece" and the first two words are the same, the two sentences are similar.
```

**Example 3:**

```
**Input:** sentence1 = ["I","love","leetcode"], sentence2 = ["I","love","onepiece"], similarPairs = [["manga","hunterXhunter"],["platform","anime"],["leetcode","platform"],["anime","manga"]]
**Output:** false
**Explanation:** "leetcode" is not similar to "onepiece".
```

**Constraints:**

* `1 <= sentence1.length, sentence2.length <= 1000`
* `1 <= sentence1[i].length, sentence2[i].length <= 20`
* `sentence1[i]` and `sentence2[i]` consist of lower-case and upper-case English letters.
* `0 <= similarPairs.length <= 2000`
* `similarPairs[i].length == 2`
* `1 <= xi.length, yi.length <= 20`
* `xi` and `yi` consist of English letters.

## ac

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

        // build union find
        UnionFind uf = new UnionFind();
        for (String[] p : pairs) {
            uf.connect(p[0], p[1]);
        }

        // check sentence
        for (int i = 0; i < words1.length; i++) {
            if (words1[i].equals(words2[i])) continue; // equal, skip
            if (!uf.fathers.containsKey(words1[i]) || !uf.fathers.containsKey(words2[i])) return false; // one of them not in map
            if ( !uf.find(words1[i]).equals(uf.find(words2[i]))) return false; // not in the same group
        }

        return true;
    }
}
class UnionFind {
    Map<String, String> fathers = new HashMap<>();
    void connect(String s1, String s2) {
        fathers.putIfAbsent(s1, s1);
        fathers.putIfAbsent(s2, s2);

        String f1 = find(s1);
        String f2 = find(s2);
        if (!f1.equals(f2)) fathers.put(f1, f2);
    }
    String find(String s) {
        if (!fathers.get(s).equals(s)) {
            fathers.put(s, find(fathers.get(s)));
        }
        return fathers.get(s);
    }
}

/*
String version UnionFind. careful nullpointer exception when word not in map.
*/
```
