Hashing-1 solution

[shakthinandana]

No description provided.

[super30admin]

Grouping Anagrams Together (problem1.py)

The student did a good job. They provided a clear solution with correct complexity analysis. One minor note: in the space complexity comment, "k is the length of array to store the anagrams" might be misinterpreted. Actually, the space complexity is O(nk) because we store all the strings (each of length up to k) and also the sorted keys (each of length k). So it's better to say "where n is the number of strings and k is the maximum length of a string". Also, note that the reference solution uses a different method (prime hashing) which has O(nk) time, but in practice your solution is common in Python and works well for the constraints.

Final

VERDICT: PASS

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Isomorphic Strings (problem2.py)

Your solution is correct and efficient. You have correctly implemented the mapping checks using two dictionaries, which ensures that no two characters map to the same character and that all mappings are consistent. Your time and space complexity analysis is accurate, and you've noted that the space is effectively constant due to the limited character set.

Strengths:

• Clear and concise code.
• Good use of zip for iteration.
• Proper handling of both forward and backward mappings.

Areas for improvement:

• The length check at the beginning is not strictly necessary given the problem constraints (t.length == s.length), but it is a good defensive practice. However, you could omit it without affecting correctness.
• Consider adding more comments to explain the logic for clarity, especially for someone new to the problem.

Overall, this is a solid solution that meets all requirements.

VERDICT: PASS

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Word Pattern (problem3.py)

Your solution is well-structured and efficient. Here are a few points for improvement:

1. Variable Naming: Consider using more descriptive names for variables. For example, wset could be renamed to mapped_words or used_words to make it clearer that it stores words that have already been mapped to a pattern character.
2. Explanation: Your time complexity analysis is slightly off. The split operation takes O(len(s)) time, which is linear in the length of the string. The loop runs O(n) times, where n is the length of the pattern (which equals the number of words). So the total time complexity is O(len(s) + n), which is acceptable. Similarly, space complexity is O(n) for the map and set, plus O(m) for the list of words (which is also O(n)), so it's O(n).
3. Edge Cases: Your solution correctly handles the case where the pattern length and number of words differ. However, consider what happens if the pattern is empty or the string is empty. The constraints say both have at least length 1, so it's safe.
4. Alternative Approach: While your solution is correct, using two dictionaries (one for pattern->word and one for word->pattern) might make the code more symmetric and easier to understand. For example:
   • If you have char_to_word and word_to_char, then you can check both mappings in a symmetric way. This is how the reference solution does it.
   • However, your current approach with a set is also valid and efficient.

Overall, good job! Your code is concise and correct.

VERDICT: PASS