737. Sentence Similarity II (Python)

Related Topic

Depth-First-Search. Graph.

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, 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”].

Constraints:

• 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]

Methodology

Build up a graph for similar word pairs. To connect a word to all its similar word, we use dictionary of set. Then iterate the two words w1 and w2 from words1 and words2 by same index. We find all similar words of w2 and their similar words recursively until we find the a word that same as w1 that means we find a transition from w2 to w1 so we return true for this pair and continue to next iteration.

Code

def areSentencesSimilarTwo(self, words1: List[str], words2: List[str], pairs: List[List[str]]) -> bool:
        if len(words1) != len(words2): return False
        similars = collections.defaultdict(set)
        for w1, w2 in pairs:
            similars[w1].add(w2)
            similars[w2].add(w1)
        def dfs(words1, words2, visited):
            for similar in similars[words2]:
                if similar in visited: continue
                if words1 == similar:
                    return True
                else:
                    visited.add(similar)
                    if dfs(words1, similar, visited):
                        return True
            return False

        for w1, w2 in zip(words1, words2):
            if w1 != w2 and not dfs(w1, w2, set([w2])):
                return False
        return True

BigO

Time Complexity: O(NP), where NN is the maximum length of words1 and words2, and PP is the length of pairs. Each of NN searches could search the entire graph.