734. Sentence Similarity - Explanation

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Description

We can represent a sentence as an array of words, for example, the sentence "I am happy with neetcode" can be represented as arr = ["I","am","happy","with","neetcode"].

Given two sentences sentence1 and sentence2 each represented as a string array and given an array of string pairs similarPairs where similarPairs[i] = [x_i, y_i] indicates that the two words x_i and y_i 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 not transitive. For example, if the words a and b are similar, and the words b and c are similar, a and c are not necessarily similar.

Example 1:

Input: sentence1 = ["great","acting","skills"], sentence2 = ["fine","drama","talent"], similarPairs = [["great","fine"],["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 = ["great"], sentence2 = ["great"], similarPairs = []

Output: true

Explanation: A word is similar to itself.

Example 3:

Input: sentence1 = ["great"], sentence2 = ["doubleplus","good"], similarPairs = [["great","doubleplus"]]

Output: false

Explanation: As they don't have the same length, we return false.

Constraints:

• 1 <= sentence1.length, sentence2.length <= 1000
• 1 <= sentence1[i].length, sentence2[i].length <= 20
• sentence1[i] and sentence2[i] consist of English letters.
• 0 <= similarPairs.length <= 1000
• similarPairs[i].length == 2
• 1 <= x_i.length, y_i.length <= 20
• x_i and y_i consist of lower-case and upper-case English letters.
• All the pairs (x_i, y_i) are distinct.

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1. Using Hash Map and Hash Set

class Solution(object):
    def areSentencesSimilar(self, sentence1, sentence2, similarPairs):
        if len(sentence1) != len(sentence2):
            return False
        
        wordToSimilarWords = defaultdict(set)

        for word1, word2 in similarPairs:
            wordToSimilarWords[word1].add(word2)
            wordToSimilarWords[word2].add(word1)

        for i in range(len(sentence1)):
            if sentence1[i] == sentence2[i] or sentence2[i] in wordToSimilarWords[sentence1[i]]:
                continue
            return False
        
        return True

Time & Space Complexity

• Time complexity: $O((n + k) \cdot m)$
• Space complexity: $O(k\cdot m)$

Where $n$ is the number of words in sentence1 and sentence2, $k$ is the length of similarPairs, and $m$ is the average length of words in sentence1 as well as similarPairs.