Hashing1

[MeghaN28]

No description provided.

[super30admin]

Grouping Anagrams Together (GroupAnagrams.java)

Your solution is correct and well-implemented. You have correctly identified the time and space complexity, and your code is clean and readable.

One area for improvement is to consider the efficiency for larger strings. The sorting approach (O(K log K) per string) might be slower than the prime product approach (O(K) per string) when the strings are very long. The prime product approach uses multiplication of prime numbers assigned to each letter, which ensures that anagrams produce the same product due to the uniqueness of prime factorization. However, note that for very long strings, the product might become very large and could lead to integer overflow in Java. You could use a double or BigInteger to mitigate this, but that might introduce other issues.

Alternatively, you could use a frequency array (count of each character) and convert that to a string to use as a key. For example, for each string, create an array of 26 integers (for each letter), count the frequency of each character, and then convert that array to a string (e.g., "#1#2#0..." ) to use as a key. This would be O(N * K) time and O(N * K) space as well, but might be more efficient than sorting for large K? Actually, K is only up to 100, so both methods are acceptable.

Overall, your solution is good and passes all test cases on LeetCode. Keep up the good work!

VERDICT: PASS

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Isomorphic Strings (IsomorphicString.java)

Your solution is excellent and matches the reference solution closely. Here are some strengths and minor suggestions:

Strengths:

• You correctly used two hash maps to maintain bidirectional mappings, which is crucial for solving the problem.
• The code is clean, well-commented, and follows good naming conventions.
• You handled edge cases implicitly by checking mappings during iteration.

Suggestions:

1. While your comment mentions "lowercase letters", the problem states that the strings consist of "any valid ascii character", which includes more than 26 characters. However, since you used HashMap, it will handle all ASCII characters correctly. The comment could be updated to reflect that the space complexity is O(1) because the number of unique characters is bounded by the ASCII set (128 or 256 characters), not just 26.
2. Consider adding a check at the beginning for the length of s and t. Although the problem states they are of equal length, it's a good practice to validate inputs if constraints aren't guaranteed (but here they are).
3. You can use Character objects without issues, but note that your solution is efficient as is.

Overall, great job! Your solution is correct and efficient.

VERDICT: PASS

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Word Pattern (Word Pattern.java)

Your solution is well-implemented and follows a similar approach to the reference solution. Here are a few points to consider for improvement:

1. Initial Check: You correctly check if the number of words matches the pattern length. This is important and efficient.
2. HashMap Usage: You use two hashmaps effectively to maintain the mappings. This ensures that both directions of the bijection are checked.
3. String Comparison: You use equals for string comparison, which is correct. Note that using == for string comparison would be incorrect as it compares references, not content.
4. Edge Cases: Your solution handles edge cases such as different lengths and mismatched mappings correctly.
5. Code Comments: Your comments are helpful, but you could make them more concise as per the problem's request (three sentences only). The current comments are slightly more verbose but still clear.

One minor optimization: You could avoid splitting the entire string if the pattern length does not match the number of words. However, since the split operation is O(n) and you already have the length check, it is efficient enough.

Overall, your solution is correct, efficient, and follows good coding practices. Keep up the good work!

VERDICT: PASS