Sentence Similarity III

#613 Most AskedMedium

Sentence Similarity III #613 Most Asked

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You are given two strings sentence1 and sentence2, each representing a sentence composed of words. A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each word consists of only uppercase and lowercase English characters.

Two sentences s1 and s2 are considered similar if it is possible to insert an arbitrary sentence (possibly empty) inside one of these sentences such that the two sentences become equal. Note that the inserted sentence must be separated from existing words by spaces.

For example,

s1 = "Hello Jane" and s2 = "Hello my name is Jane" can be made equal by inserting "my name is" between "Hello" and "Jane" in s1.
s1 = "Frog cool" and s2 = "Frogs are cool" are not similar, since although there is a sentence "s are" inserted into s1, it is not separated from "Frog" by a space.

Given two sentences sentence1 and sentence2, return true if sentence1 and sentence2 are similar. Otherwise, return false.

Example 1:

Input: sentence1 = "My name is Haley", sentence2 = "My Haley"

Output: true

Explanation:

sentence2 can be turned to sentence1 by inserting "name is" between "My" and "Haley".

Example 2:

Input: sentence1 = "of", sentence2 = "A lot of words"

Output: false

Explanation:

No single sentence can be inserted inside one of the sentences to make it equal to the other.

Example 3:

Input: sentence1 = "Eating right now", sentence2 = "Eating"

Output: true

Explanation:

sentence2 can be turned to sentence1 by inserting "right now" at the end of the sentence.

Constraints:

1 <= sentence1.length, sentence2.length <= 100
sentence1 and sentence2 consist of lowercase and uppercase English letters and spaces.
The words in sentence1 and sentence2 are separated by a single space.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can you provide an upper bound on the length of the input sentences (sentence1 and sentence2)?
Are the input sentences case-sensitive, or should I consider them case-insensitive?
What should be returned if the sentences are identical?
Are there any special characters or punctuation marks present in the sentences besides spaces?
If multiple possible longest common prefixes exist, should I return the length of any of them, or is there a specific criteria for choosing one?

Brute Force Solution

Approach

The brute force strategy involves trying out every possible combination to find the shortest sequence of operations to make two sentences similar. This means checking every possible position to insert or delete words to make the sentences identical.

Here's how the algorithm would work step-by-step:

Try aligning the first sentence with the second sentence as they are.
Then, try inserting a word from the second sentence into the first sentence at every possible position.
Next, try inserting two words, then three words, and so on, at every possible position.
Do the same process of inserting words from the first sentence into the second sentence.
For each combination of insertions, see if the resulting sentences are identical.
Keep track of the number of insertions made in each successful attempt.
Return the smallest number of insertions needed to make the sentences the same.

Code Implementation

def sentence_similarity_brute_force(sentence1, sentence2):
    words1 = sentence1.split()
    words2 = sentence2.split()

    minimum_insertions_needed = float('inf')

    # Helper function to check if two lists of words are equal
    def are_sentences_equal(sentence1_words, sentence2_words):
        return sentence1_words == sentence2_words

    # Iterate through all possible numbers of insertions
    for number_of_insertions in range(len(words2) + 1):
        # Iterate through all possible positions to insert
        for i in range(len(words1) + 1):
            # Create a new sentence with the insertion
            new_sentence = words1[:i] + words2[:number_of_insertions] + words1[i:]

            # Check if sentences are now equal
            if are_sentences_equal(new_sentence, words2):
                minimum_insertions_needed = min(minimum_insertions_needed, number_of_insertions)

    for number_of_insertions in range(len(words1) + 1):
        for i in range(len(words2) + 1):
            new_sentence = words2[:i] + words1[:number_of_insertions] + words2[i:]

            if are_sentences_equal(new_sentence, words1):
                minimum_insertions_needed = min(minimum_insertions_needed, number_of_insertions)

    # Handle case where the original sentences are equal
    if are_sentences_equal(words1, words2):
        return 0

    # If no combination resulted in the same sentence,
    # return 'inf'
    if minimum_insertions_needed == float('inf'):
        return -1

    return minimum_insertions_needed

Big(O) Analysis

Time Complexity

O(n!) – The brute force strategy explores all possible insertion combinations to transform one sentence into the other. In the worst case, we might need to insert words from one sentence into the other at every possible position, considering all possible subsequences. If the length of the longer sentence is 'n', the number of possible insertions and their arrangements grows factorially. Therefore, the time complexity is proportional to the number of these combinations, which is O(n!).

Space Complexity

O(N^2) – The described brute force approach explores all possible insertions. In the worst-case scenario, where sentences are very different, we might be creating modified sentences by inserting words at every possible position. Creating these modified sentences can require storing up to N words, and trying all combinations of insertions leads to an auxiliary space complexity that is, in the worst case, proportional to the number of possible sentence combinations, or N^2, where N is the total number of words across both sentences. Temporary strings, lists, or other data structures are needed to store the intermediate modified sentences, contributing to the O(N^2) space complexity. Specifically, the algorithm might create copies of at least one of the sentences with insertions at multiple locations and for multiple insertion lengths which are stored in memory at a given instant.

Optimal Solution

Approach

The most efficient method to determine sentence similarity involves comparing the sentences from both ends inwards. If they are similar, one sentence can be transformed into the other by inserting words, and this transformation happens at one point, either at the beginning, the end, or somewhere in the middle.

Here's how the algorithm would work step-by-step:

Begin by comparing the sentences from the start until you find a difference.
Also, compare the sentences from the end until you find a difference.
If there's a part of the shorter sentence that matches both the beginning and end of the longer sentence (after accounting for differences), it means the remaining words from the longer sentence can be inserted into the shorter one.
Specifically, remove the matching beginning and ending parts from the longer sentence.
If what's left of the longer sentence can be empty, it means the sentences are similar because we can insert the extra words from the longer sentence into the shorter one to make them identical. Return true in this case, otherwise return false.

Code Implementation

def areSentencesSimilar(sentence1, sentence2):
    words1 = sentence1.split()
    words2 = sentence2.split()
    
    index_left = 0
    index_right1 = len(words1) - 1
    index_right2 = len(words2) - 1

    # Find matching words from the beginning
    while index_left <= index_right1 and index_left <= index_right2 and words1[index_left] == words2[index_left]:
        index_left += 1

    # Find matching words from the end
    while index_right1 >= index_left and index_right2 >= index_left and words1[index_right1] == words2[index_right2]:
        index_right1 -= 1
        index_right2 -= 1

    # Check if the shorter sentence's unmatched part is empty
    if len(words1) > len(words2):
        # Determine if words2 can be inserted into words1
        return index_right2 < index_left
    else:
        # Determine if words1 can be inserted into words2
        return index_right1 < index_left

Big(O) Analysis

Time Complexity

O(n) – The algorithm compares the two sentences from the beginning and the end simultaneously. The comparisons continue until a mismatch is found in both directions. The number of comparisons in each direction is bounded by the length of the shorter sentence. Therefore, the time complexity is proportional to the length of the shorter sentence, which we can represent as n. Thus, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm primarily creates lists from the input sentences using the split() method, resulting in two lists whose combined length is proportional to the total number of words in both sentences. Let N represent the total number of words across the two sentences. The split() operation consumes space proportional to N to store the word lists. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Both sentences are empty strings	Return True because an empty sentence is similar to another empty sentence.
One sentence is empty, and the other isn't	Return True because the empty sentence can become the non-empty sentence through insertion of all words from the latter.
Sentences are identical	Return True immediately since identical sentences are trivially similar.
One sentence is a prefix or suffix of the other.	The algorithm should correctly identify the insertions needed to transform the shorter string into the longer one.
Sentences have some overlapping words, but neither is a prefix or suffix of the other.	The algorithm should explore both prepending and appending words to the shorter string, selecting the path with the minimal number of insertions required.
Sentences are very long (e.g., 1000 words each).	Ensure the solution uses an efficient algorithm (e.g., dynamic programming) to avoid exceeding time limits and memory constraints due to quadratic complexity.
Sentences contain the same words but in different orders and with no common prefix or suffix.	Return False as no simple insertion or deletion can resolve to transform them.
Sentences have only one word each.	Compare the words and return True if they are the same, and False otherwise.

Both sentences are empty strings

How to Handle:

Return True because an empty sentence is similar to another empty sentence.

One sentence is empty, and the other isn't

How to Handle:

Return True because the empty sentence can become the non-empty sentence through insertion of all words from the latter.

Sentences are identical

How to Handle:

Return True immediately since identical sentences are trivially similar.

One sentence is a prefix or suffix of the other.

How to Handle:

The algorithm should correctly identify the insertions needed to transform the shorter string into the longer one.

Sentences have some overlapping words, but neither is a prefix or suffix of the other.

How to Handle:

The algorithm should explore both prepending and appending words to the shorter string, selecting the path with the minimal number of insertions required.

Sentences are very long (e.g., 1000 words each).

How to Handle:

Ensure the solution uses an efficient algorithm (e.g., dynamic programming) to avoid exceeding time limits and memory constraints due to quadratic complexity.

Sentences contain the same words but in different orders and with no common prefix or suffix.

How to Handle:

Return False as no simple insertion or deletion can resolve to transform them.

Sentences have only one word each.

How to Handle:

Compare the words and return True if they are the same, and False otherwise.