737. Sentence Similarity II

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.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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.

Solution

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
class Solution {
    int n = 0;
    int[] uf = new int[10_000];
    int[] sizes = new int[10_000];
    Map<String, Integer> indexes = new HashMap<>();
    
    public boolean areSentencesSimilarTwo(String[] sen1, String[] sen2, List<List<String>> similarPairs) {
        
        if (sen1.length != sen2.length) return false;

        for (List<String> pair: similarPairs) {
            add(pair.get(0), pair.get(1));
        }

        for (int i = 0; i < sen1.length; i++) {

            var word1 = sen1[i];
            var word2 = sen2[i];

            if (word1.equals(word2)) {
                continue;
            }

            if (!contains(word1)) {
                return false;
            }

            if (!contains(word2)) {
                return false;
            }

            if (!isConnected(word1, word2)) {
                return false;
            }
        }

        return true;
    }
            
    void add(String w1, String w2) {
        if (!indexes.containsKey(w1)) {
            indexes.put(w1, n);
            uf[n] = n;
            sizes[n] = 1;
            n++;
        }
        
        if (!indexes.containsKey(w2)) {
            indexes.put(w2, n);
            uf[n] = n;
            sizes[n] = 1;
            n++;
        } 
        
        int pid = parent(indexes.get(w1));
        int qid = parent(indexes.get(w2));
        
        if (sizes[pid] > sizes[qid]) {
            uf[qid] = pid;
            sizes[pid] += sizes[qid];
        } else {
            uf[pid] = qid;
            sizes[qid] += sizes[pid];
        }
    }
            
    boolean contains(String word) {
        return indexes.containsKey(word);
    }       
            
    boolean isConnected(String w1, String w2) {
        return parent(indexes.get(w1)) == parent(indexes.get(w2));
    } 
            
    int parent(int pid) {
        while (pid != uf[pid]) {
            pid = uf[pid];
        }
        
        return pid;
    }
}

Solution#

Solution