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

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.
*/