Problem Description

The problem asks to determine if two sentences are similar. Each sentence is represented as an array of words. For example, the sentence "I am happy with leetcode" can be represented as arr = ["I", "am", "happy", "with", "leetcode"]. The words within each sentence are considered similar based on a provided list of word pairs similarPairs, where each pair [xi, yi] indicates that xi and yi are similar words. Two sentences are similar if they have the same number of words and each word in a position in sentence1 is similar to the corresponding word in the same position in sentence2. A word is always considered similar to itself. It is important to note that the similarity between words is not transitive - just because word a is similar to word b and word b is similar to word c, it doesn't mean a is similar to c.

Intuition

To determine if the two sentences are similar, there are two main checks that need to be carried out:

1. Check Length: First, we need to check if both sentences have the same number of words. If they don't, we can immediately return False as they can't be similar with different lengths.

2. Check Word Similarity: Then, we check each pair of corresponding words from sentence1 and sentence2. For each word pair (a, b) we check:

   • If a is the same as b, they are automatically similar.
   • If a and b are found as a pair in similarPairs, they are similar.
   • As similarity is not directed (it is bidirectional), we should also check (b, a) in similarPairs.

If all word pairs pass these similarity checks, the sentences are similar, and we return True. Otherwise, at the first instance of dissimilarity, the overall similarity check fails, and we return False.

The solution uses set comprehension to build a set s from similarPairs for quick lookup, and then utilizes the all function combined with list comprehension to efficiently carry out the similarity checks for each word pair from the zipped sentences.

Solution Approach

The implemented solution takes advantage of the following algorithms, data structures, and programming constructs:

• Set Data Structure: A set is used to store the similarPairs for efficient O(1) average time complexity on lookups. This is done by creating a set s using set comprehension to hold all given similar word pairs in both directions (e.g., (a, b) and (b, a)). The nature of a set provides constant time membership testing, which is much faster compared to a list or array for this purpose.

• Zip Function: The zip function is a built-in Python function that aggregates elements from two or more iterables (arrays, lists, etc.) and returns an iterator of tuples. In this solution, zip is used to create pairs of corresponding words from sentence1 and sentence2.

• List Comprehension with all Function: A list comprehension is used in conjunction with the all function to check if all corresponding pairs of words between sentence1 and sentence2 are similar. The all function returns True if all elements of the given iterable are true (or if the iterable is empty). The list comprehension iterates through each word pair and uses a logical or || to evaluate if the words are identical, or if they exist as similar word pairs in the set s.

Here is the step-by-step approach of the solution process:

1. Check Length Equality:

   • Before iterating through the words, the solution checks if sentence1 and sentence2 have the same length using len(sentence1) != len(sentence2). If they don't match, it returns False, thereby optimizing the solution by eliminating non-matching sentences early on.

2. Iterative Comparison with Early Exit:

   • The solution iterates over each pair of words (a, b) from the zipped sentences using list comprehension and checks:
     • If a == b, then the words are the same.
     • If (a, b) in s, the pair exists in the similarity set, so they are similar.
     • If (b, a) in s, since similarity is not transitive but is bidirectional, this pair is also checked in case the reverse was provided in similarPairs.
   • The all function wraps the list comprehension to ensure that every pair meets at least one of these conditions. If even one pair does not meet the similarity condition, all returns False.

By using this approach, the solution effectively breaks down the problem into small logical checks, ensuring an efficient and clear way to compare the sentences. This results in a concise and effective code:

def areSentencesSimilar(self, sentence1: List[str], sentence2: List[str], similarPairs: List[List[str]]) -> bool:
    if len(sentence1) != len(sentence2):
        return False
    s = {(a, b) for a, b in similarPairs}
    return all(
        a == b or (a, b) in s or (b, a) in s for a, b in zip(sentence1, sentence2)
    )

Example Walkthrough

Let's consider a small example to illustrate the solution approach.

Suppose we have two sentences represented by the arrays below:

• sentence1 = ["I", "love", "to", "code"]
• sentence2 = ["I", "adore", "to", "program"]

And let's say we are also given the list of similar word pairs:

• similarPairs = [["love", "adore"], ["code", "program"]]

When we apply our solution, this is how it will proceed:

1. Check Length Equality:

   • Both sentence1 and sentence2 have the same number of words (4 words each), so we move on to the next step. If they had a different number of words, we would return False at this point.

2. Iterative Comparison with Early Exit:

   • We create an iterator of tuples by pairing up corresponding items from both sentences using zip(sentence1, sentence2), which gives us [("I", "I"), ("love", "adore"), ("to", "to"), ("code", "program")].
   • We create a set from similarPairs to store all given similar word pairs, including their bidirectional counterparts - that would give us a set s containing [("love", "adore"), ("adore", "love"), ("code", "program"), ("program", "code")].
   • Using list comprehension, we check each pair:
     • First pair is ("I", "I"): the words are identical, so this pair is similar.
     • Second pair is ("love", "adore"): the pair exists in set s, so this pair is also similar.
     • Third pair is ("to", "to"): again, the words are identical, similar by default.
     • Fourth pair is ("code", "program"): the pair exists in set s, hence this pair is similar too.
   • Since every pair of words from both sentences have passed the similarity check, the all function will ultimately return True.

Using this approach with the given sentence1, sentence2, and similarPairs, our function areSentencesSimilar would return True because all corresponding word pairs are similar as per our rules.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code can be broken down as follows:

1. Checking if len(sentence1) is equal to len(sentence2) takes O(1) time, as it is simply comparing two integers.
2. Creating a set s from the similarPairs takes O(p) time, where p is the number of similar pairs.
3. The list comprehension iterates over each pair of words from sentence1 and sentence2, which occurs in O(n) time, where n is the number of words in each sentence (since we've previously established the sentences are of equal length to reach this point).
4. Each comparison within the list comprehension is O(1) because checking equality of strings and checking for existence in a set are both constant-time operations.

Thus, the time complexity is O(n + p), where n is the length of the sentences and p is the number of similar word pairs.

Space Complexity

The space complexity can be analyzed as follows:

1. The additional set s which stores the similar word pairs requires O(p) space, where p is the number of similar pairs.
2. The space for input variables sentence1, sentence2, and similarPairs is not counted as extra space as per convention since they are part of the input.

As a result, the auxiliary space complexity of the code is O(p).

Learn more about how to find time and space complexity quickly using problem constraints.