734. Sentence Similarity

Description

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].

Solution

HashMap, O(m + n), S(n)

m: words1.length(), should be same with words2.length()
n: pairs.length

题目很清晰地交代了几点：

• 如果s1和s2相似，那么s2和s1必然相似；
• 如果s1和s2相似，s2和s3相似，那么s1和s3不一定相似；(这点还挺关键的，一定要弄清楚)
• s1和自己相似，无论pairs中包不包含s1

那么就可以用最直观的HashMap解法。

class Solution {
    public boolean areSentencesSimilar(
        String[] words1, String[] words2, String[][] pairs) {
        
        if (words1 == null || words2 == null || pairs == null 
            || words1.length != words2.length) {
            return false;
        }
        
        Map> map = new HashMap<>();
        for (String[] pair : pairs) {
            add(pair[0], pair[1], map);
            add(pair[1], pair[0], map);
        }
        
        for (int i = 0; i < words1.length; ++i) {
            String s1 = words1[i];
            String s2 = words2[i];
            if (!s1.equals(s2) 
                && !map.getOrDefault(s1, Collections.EMPTY_SET).contains(s2)) {
                return false;
            }
        }
        
        return true;
    }
    
    private void add(String word, String similar, Map> map) {
        if (!map.containsKey(word)) {
            map.put(word, new HashSet<>());
        }
        map.get(word).add(similar);
    }
}