737. Sentence Similarity II
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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, words1 = ["great", "acting", "skills"]
and words2 = ["fine", "drama", "talent"]
are similar, if the similar word pairs are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]]
.
Note that the similarity relation is transitive. For example, if "great" and "good" are similar, and "fine" and "good" are similar, then "great" and "fine" are similar.
Similarity is also 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:
words1
and words2
will not exceed 1000
.pairs
will not exceed 2000
.pairs[i]
will be 2
.words[i]
and pairs[i][j]
will be in the range [1, 20]
.class Solution: '''union-find''' def areSentencesSimilarTwo(self, words1, words2, pairs): """ :type words1: List[str] :type words2: List[str] :type pairs: List[List[str]] :rtype: bool """ if len(words1)!=len(words2): return False d, cnt = {}, 0 for pair in pairs: for i in pair: if i not in d: d[i] = cnt cnt += 1 adj = list(range(len(d))) def union(i, j): f1, f2 = find(i), find(j) adj[f1] = f2 def find(i): while adj[i]!=i: i=adj[i] return i for i, j in pairs: union(d[i], d[j]) for i in range(len(words1)): if words1[i]==words2[i]:continue if words1[i] in d and words2[i] in d and find(d[words1[i]])==find(d[words2[i]]):continue return False return True s = Solution()print(s.areSentencesSimilarTwo(["great", "acting", "skills"], ["fine", "drama", "talent"], [["great", "fine"], ["acting","drama"], ["skills","talent"]]))
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