hashing-1 solutions

[ankurkhanna0405]

No description provided.

[super30admin]

Grouping Anagrams Together (groupAnagrams.py)

• The solution is correct and handles all cases.
• Time complexity is O(N * K log K), which is acceptable but not optimal. The reference solution has O(N * K) which is better. However, in Python, the sorting method is often used because it is straightforward and efficient in practice for typical input sizes.
• Space complexity is O(N * K), which is necessary for storing the groups. However, creating a sorted string for each input might use extra space. Alternatively, we could use a tuple of the sorted characters as the key without joining into a string? But that might not save much.
• Code quality: Use more descriptive variable names (e.g., anagram_groups instead of mapp). Also, iterate directly over the elements of strs instead of using indices.
• Suggestion: The student can consider using defaultdict(list) to avoid the if-else check.

Improved version without changing the algorithm:

from collections import defaultdict

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        anagram_groups = defaultdict(list)
        for s in strs:
            key = ''.join(sorted(s))
            anagram_groups[key].append(s)
        return list(anagram_groups.values())

This is more Pythonic.

VERDICT: PASS

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Isomorphic Strings (isIsomorphic.py)

Your solution is correct and efficient. You have correctly implemented the necessary checks to ensure that the mapping is consistent and that no two characters in s map to the same character in t. However, there are a few improvements you can make to enhance code clarity and Pythonic style:

1. Variable Names: Use more descriptive names. For example, instead of mapp, use s_to_t_mapping or simply mapping. Instead of sett, use mapped_t_chars or used_t_chars.

2. Iteration: Instead of iterating by index, consider using zip(s, t) to iterate over pairs of characters. This is more Pythonic and can make the code cleaner.

3. Edge Cases: Although the problem states that the strings are of equal length, it's good practice to add a check at the beginning to return false if the lengths are different. However, given the constraint, this is optional.

Here's a revised version of your code with these improvements:

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        mapping = {}
        mapped_t_chars = set()
        for char_s, char_t in zip(s, t):
            if char_s in mapping:
                if mapping[char_s] != char_t:
                    return False
            else:
                if char_t in mapped_t_chars:
                    return False
                mapping[char_s] = char_t
                mapped_t_chars.add(char_t)
        return True

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

VERDICT: PASS

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Word Pattern (wordPattern.py)

Your solution is well-structured and follows the problem requirements correctly. Here are some strengths and areas for improvement:

Strengths:

• You correctly handle the bijection requirement by using both a dictionary (for pattern-to-word mapping) and a set (to track used words). This ensures that no two pattern characters map to the same word and no two words map to the same pattern character.
• The initial check for length mismatch is efficient and necessary.
• The code is readable and logically organized.

Areas for Improvement:

• While your solution is correct, note that the reference solution uses two dictionaries (or maps) to maintain both mappings. Your approach with one dictionary and one set is equally efficient and achieves the same goal. However, using two dictionaries might make the symmetry more explicit. Both approaches are valid.
• Consider adding a comment to explain why you are using the set (to ensure words are not reused) for clarity.
• The variable name link could be more descriptive, such as char_to_word to indicate its purpose.
• There is a minor indentation issue in the else block: the line link[pattern[i]]=words[i] should be indented consistently. However, this might be a formatting issue in the provided code snippet.

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

VERDICT: PASS