Complete Hashing-1

[satish-paraddi]

No description provided.

[super30admin]

Group Anagrams (Problem-1.py)

Strengths:

1. Your solution correctly implements the prime product hashing technique, which is an elegant approach for grouping anagrams.
2. Good code organization with the helper method extracted.
3. Helpful comments at the top documenting time/space complexity and approach.
4. The solution runs successfully on LeetCode.

Areas for improvement:

1. Variable naming: Consider using more descriptive names like anagram_groups instead of d for better readability.
2. Use defaultdict: You can simplify the code by using collections.defaultdict(list) to avoid the if/else check:

   from collections import defaultdict
   d = defaultdict(list)
   for s in strs:
       d[self.primeProduct(s)].append(s)
   return list(d.values())

3. Prime list: Remove the extra prime (103) since you only need 26 primes for lowercase English letters.
4. Indentation: The indentation in the primeProduct method is inconsistent. Make sure to use consistent 4-space indentation throughout.
5. Potential overflow consideration: While Python handles big integers natively, in other languages prime products can overflow. Consider mentioning this or using a different approach (like character count tuple) for robustness.

VERDICT: PASS

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Isomorphic Strings (Problem-2.py)

Strengths:

1. Your solution correctly implements the bidirectional mapping check using a dictionary and a set.
2. The time and space complexity match the reference solution.
3. The code is clean and readable.
4. Good use of comments at the top.

Areas for Improvement:

1. Variable Naming: Consider using more descriptive variable names like s_to_t_map instead of prim, and used_chars instead of second. This would make the code more self-documenting.
2. Alternative Approach: You could also use two dictionaries (like the reference solution) for symmetry, though your approach with a set is equally valid and slightly more memory-efficient.
3. Edge Case Documentation: While the problem constraints guarantee equal length, adding a comment about this assumption could be helpful.

VERDICT: PASS

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Word Pattern (Problem-3.py)

Strengths:

• Clean, concise solution that correctly handles the bijection requirement
• Good use of early returns for efficiency
• Helpful header comments documenting complexity and approach

Areas for Improvement:

1. Time Complexity Claim: Your comment states O(n+m), but the s[i] in h.values() check makes it O(n×m) in the worst case. Consider using two hash maps for true O(n) complexity:

   char_to_word = {}
   word_to_char = {}
   for i in range(len(p)):
       if p[i] not in char_to_word:
           char_to_word[p[i]] = s[i]
       elif char_to_word[p[i]] != s[i]:
           return False
       if s[i] not in word_to_char:
           word_to_char[s[i]] = p[i]
       elif word_to_char[s[i]] != p[i]:
           return False

2. Variable Naming: Reassigning s from a string to a list is confusing. Consider using different variable names like words = s.split(' ') to avoid shadowing the input parameter.

3. Edge Case Consideration: Your solution handles the length mismatch case well, but consider what happens with empty strings or single characters (though constraints prevent these).

VERDICT: PASS