Problem Description

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 are considered similar if you can insert an arbitrary sentence (possibly empty) inside one of these sentences to make them equal. The key requirement is that the inserted sentence must be separated from existing words by spaces.

For example:

• sentence1 = "Hello Jane" and sentence2 = "Hello my name is Jane" are similar because you can insert "my name is" between "Hello" and "Jane" in sentence1 to make them equal.
• sentence1 = "Frog cool" and sentence2 = "Frogs are cool" are not similar. Even though you could think of inserting "s are" into sentence1, this would require modifying the word "Frog" to "Frogs", which violates the rule that inserted text must be separated by spaces.

The task is to determine if sentence1 and sentence2 are similar according to this definition. Return true if they are similar, otherwise return false.

The solution uses a two-pointer approach. After splitting both sentences into word arrays, it checks how many words match from the beginning and from the end of both sentences. If the total number of matching words from both ends covers the entire shorter sentence, then the sentences are similar (since the remaining words in the longer sentence could be the "inserted" portion).

Intuition

The key insight is that if two sentences are similar, one sentence can be obtained from the other by inserting a continuous block of words at some position. This means the beginning and/or ending portions of both sentences must match.

Think about it this way: if we can insert words into the shorter sentence to make it equal to the longer sentence, then the shorter sentence must be a "subsequence" of the longer sentence where the matching parts are either at the beginning, at the end, or split between both ends.

For instance, if we have "A B C D E" and "A B E", we can see that "A B" matches from the start and "E" matches from the end. The words "C D" in the middle of the longer sentence are what we would need to insert into the shorter sentence between "B" and "E".

This leads us to a two-pointer strategy:

1. Count how many words match from the beginning of both sentences (let's call this i)
2. Count how many words match from the end of both sentences (let's call this j)
3. If i + j >= n (where n is the length of the shorter sentence), then the sentences are similar

Why does i + j >= n work? Because if the sum of matching words from both ends covers or exceeds the total words in the shorter sentence, it means all words in the shorter sentence have a corresponding match in the longer sentence, with possibly some words in between that could be the "inserted" portion. The >= handles the case where the matching portions might overlap, which would happen when the sentences are identical or when the insertion point is at the very beginning or end.

Learn more about Two Pointers patterns.

Solution Approach

The implementation follows a two-pointer approach to check if the sentences are similar:

Step 1: Parse and Prepare

• Split both sentences into word arrays using split(): words1 and words2
• Get the lengths m and n of the two word arrays
• To simplify the logic, ensure words1 is always the longer array by swapping if necessary (if m < n, swap the arrays and their lengths)

Step 2: Match from the Beginning

• Initialize pointer i = 0
• While i < n and words1[i] == words2[i], increment i
• This counts how many words match from the start of both sentences

Step 3: Match from the End

• Initialize pointer j = 0
• While j < n and words1[m - 1 - j] == words2[n - 1 - j], increment j
• This counts how many words match from the end of both sentences
• Note: We use m - 1 - j and n - 1 - j to access elements from the end

Step 4: Check Similarity Condition

• Return i + j >= n
• If the total number of matching words from both ends covers the entire shorter sentence, then the sentences are similar
• The >= handles cases where the matching portions might overlap (when sentences are identical or when the insertion is at the very beginning or end)

Example Walkthrough:

• sentence1 = "Hello Jane", sentence2 = "Hello my name is Jane"
• After splitting: words1 = ["Hello", "Jane"], words2 = ["Hello", "my", "name", "is", "Jane"]
• After ensuring longer array: words1 = ["Hello", "my", "name", "is", "Jane"], words2 = ["Hello", "Jane"]
• Matching from start: i = 1 (only "Hello" matches)
• Matching from end: j = 1 (only "Jane" matches)
• Check: i + j = 2 >= 2 (length of shorter sentence), so return true

Example Walkthrough

Let's walk through a concrete example to understand how the solution works.

Example:

• sentence1 = "My name is Haley"
• sentence2 = "My Haley"

Step 1: Parse and Prepare

• Split into words:
  • words1 = ["My", "name", "is", "Haley"] (length m = 4)
  • words2 = ["My", "Haley"] (length n = 2)
• Since m > n, no swap needed (words1 remains the longer array)

Step 2: Match from the Beginning

• Start with i = 0
• Compare words1[0] with words2[0]: "My" == "My" ✓
• Increment i to 1
• Compare words1[1] with words2[1]: "name" != "Haley" ✗
• Stop here, so i = 1

Step 3: Match from the End

• Start with j = 0
• Compare words1[3] with words2[1]: "Haley" == "Haley" ✓
• Increment j to 1
• Compare words1[2] with words2[0]: "is" != "My" ✗
• Stop here, so j = 1

Step 4: Check Similarity

• Total matches: i + j = 1 + 1 = 2
• Length of shorter sentence: n = 2
• Since 2 >= 2, return true

This makes sense because we can insert "name is" between "My" and "Haley" in sentence2 to get sentence1. The algorithm correctly identifies that all words in the shorter sentence have matching positions in the longer sentence.

Solution Implementation

Time and Space Complexity

The time complexity is O(L), where L is the sum of the lengths of the two sentences. This is because:

• Splitting both sentences takes O(L) time as we need to traverse through all characters to identify word boundaries
• The two while loops together iterate through at most n words (the length of the shorter word list), and comparing words involves character-by-character comparison
• In the worst case, we compare all characters in the sentences, which is bounded by O(L)

The space complexity is O(L), where L is the sum of the lengths of the two sentences. This is because:

• The split() operation creates two new lists words1 and words2 that store all the words from both sentences
• The total space needed to store these word lists is proportional to the total number of characters in both sentences
• The additional variables m, n, i, and j only use O(1) extra space

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

Common Pitfalls

Pitfall 1: Overcounting When Prefix and Suffix Overlap

The Problem: When the prefix and suffix matching regions overlap, we might count the same words twice. For example, if sentence1 = "A B C" and sentence2 = "A B C", the prefix match would count all 3 words, and the suffix match would also count all 3 words, giving us prefix_match_count + suffix_match_count = 6, which is greater than the actual length of 3.

Why This Isn't Actually a Problem: The current solution handles this correctly by checking prefix_match_count + suffix_match_count >= len2. The >= operator ensures that even if we overcount due to overlap, the condition still evaluates correctly. When sentences are identical or when the insertion point is at the very beginning or end, overlap is expected and handled properly.

Pitfall 2: Not Handling Empty Sentences

The Problem: If either sentence is empty or contains only whitespace, the split operation might produce unexpected results or the algorithm might not handle edge cases correctly.

Solution: Add explicit checks for empty or whitespace-only strings:

def areSentencesSimilar(self, sentence1: str, sentence2: str) -> bool:
    # Handle edge cases with empty or whitespace strings
    if not sentence1.strip() or not sentence2.strip():
        return sentence1.strip() == sentence2.strip()
  
    words1 = sentence1.split()
    words2 = sentence2.split()
    # ... rest of the code remains the same

Pitfall 3: Incorrect Index Calculation When Matching from End

The Problem: A common mistake is using incorrect indices when matching from the end. For instance, using words1[len1 - j] instead of words1[len1 - 1 - j] would cause an off-by-one error, potentially leading to index out of bounds or incorrect comparisons.

Solution: Always remember that for 0-indexed arrays:

• The last element is at index length - 1
• The j-th element from the end is at index length - 1 - j

# Correct way to access from the end
while suffix_match_count < len2 and words1[len1 - 1 - suffix_match_count] == words2[len2 - 1 - suffix_match_count]:
    suffix_match_count += 1

Pitfall 4: Not Ensuring Consistent Array Order

The Problem: If you forget to swap the arrays when len1 < len2, the algorithm logic breaks because it assumes words1 is always the longer (or equal length) array. This could lead to incorrect results or index out of bounds errors.

Solution: Always ensure the arrays are ordered consistently before processing:

# Essential swap to maintain the invariant that words1 is longer
if len1 < len2:
    words1, words2 = words2, words1
    len1, len2 = len2, len1

Pitfall 5: Misunderstanding the Problem Requirements

The Problem: A common misunderstanding is thinking that word modifications are allowed (like changing "Frog" to "Frogs"). The problem strictly requires that insertions must be complete words separated by spaces, not modifications to existing words.

Solution: The two-pointer approach naturally handles this correctly by only comparing complete words. No partial word matching or modification is performed - words must match exactly or they don't match at all.