Completed Hashing 1

[pratikb0501]

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[super30admin]

Grouping Anagrams Together (Problem_1.py)

Your solution is correct and efficient. You have successfully implemented a common approach for grouping anagrams by using character frequency counts. Here are some suggestions for improvement:

1. Variable Naming: Consider using more descriptive variable names. For example, hashCode might be better named as key or signature since it is a string representation of the frequency array. Also, my_dict could be named groups or anagram_map to clarify its purpose.

2. Code Structure: The code is well-structured, but you should avoid including test code inside the same file as the class definition in a production environment. Instead, use separate test files or unit tests.

3. Efficiency Note: While your solution is efficient, note that building a string key for each word might be slightly slower than using a tuple of the frequency array (which can be used as a key directly) or using a prime number hash (as in the reference solution). However, your method is straightforward and easy to understand.

4. Edge Cases: Your solution handles edge cases like empty strings correctly. For an empty string, the frequency array will be all zeros, and the key will be a string of 26 zeros with "#", which is correct.

5. Best Practices: You followed good practices by using a dictionary to group the anagrams. You can also use defaultdict(list) from the collections module to simplify the code a bit. For example:

   from collections import defaultdict
   my_dict = defaultdict(list)
   for word in strs:
        # compute key
        my_dict[key].append(word)

   This avoids the if-else check.

Overall, your solution is excellent. Keep up the good work!

VERDICT: PASS

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Isomorphic Strings (Problem_2.py)

Your solution is correct and efficient. You have used a dictionary to map characters from s to t and a set to ensure that no two characters in s map to the same character in t. This approach meets the problem requirements.

Strengths:

• You handle both conditions: mapping from s to t must be consistent, and no two s characters map to the same t character.
• The code is clean and easy to read.
• You included a length check which, while not necessary due to constraints, shows defensive programming.

Areas for improvement:

• The test code at the bottom (with s = "f11" and t = "b23") should not be part of the solution class. In a coding platform, only the class and method are expected. You can remove that to avoid any runtime issues in an automated judge.
• While your use of a set is correct, some might find the two-map approach (like in the reference solution) more symmetric and straightforward. However, your solution is equally valid.
• You can consider adding comments to explain the two conditions you are checking, which would make the code more understandable.

Overall, you did a great job. The solution is optimal in time and space.

VERDICT: PASS

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Word Pattern (Problem_3.py)

Your solution is well-structured and correctly solves the problem. You have correctly implemented the bijection check by using a dictionary for pattern-to-word mapping and a set to ensure words are not reused. This is efficient and clear.

Strengths:

• You correctly handle the case where the pattern length and word count differ.
• You use a set to enforce that no two pattern characters map to the same word, which is necessary for the bijection.
• The code is concise and easy to understand.

Areas for Improvement:

• Consider renaming sSet to something like usedWords to make it more descriptive and avoid confusion with the input string s.
• You might want to add a comment explaining why you check the set when adding a new pattern character, to clarify that it ensures the word isn't already mapped to a different pattern character.

Overall, your solution is correct and efficient. Good job!

VERDICT: PASS