h1

[spencerkrebs]

No description provided.

[super30admin]

Grouping Anagrams Together (group-anagrams.py)

Your solution is correct and efficient, with a time and space complexity that matches the optimal approach. Well done on implementing a working solution with clear code and comments.

One area for improvement is the key used in the dictionary. Instead of converting the character count list to a string, consider using a tuple. Tuples are hashable and can be used as dictionary keys directly. This would be more efficient and idiomatic. For example:

key = tuple(charCount)
if key not in mp:
    mp[key] = [word]
else:
    mp[key].append(word)

Additionally, you can use collections.defaultdict(list) to simplify the code. This would avoid the need to check if the key exists. For example:

from collections import defaultdict
mp = defaultdict(list)
for word in strs:
    charCount = [0]*26
    for char in word:
        charCount[ord(char)-ord('a')] += 1
    mp[tuple(charCount)].append(word)
return list(mp.values())

Another point: the problem constraints state that the strings consist of lowercase English letters, so the character count array of size 26 is appropriate.

Overall, your solution is good, but these small changes could make it more efficient and Pythonic.

VERDICT: PASS

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Isomorphic Strings (isomorphic-strings.py)

Your solution is mostly correct and efficient. You have correctly 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 satisfies the conditions of the problem.

Strengths:

• The code is clean and easy to read.
• You have considered the constraints and used efficient data structures (dictionary and set) with constant time operations.
• You provided comments explaining time and space complexity.

Areas for improvement:

1. The initial length check is redundant because the problem states that the two strings have the same length. You can remove it to make the code simpler.
2. While your solution is correct, it uses a set to track mapped t characters. Alternatively, you could use two dictionaries (one for mapping from s to t and one from t to s) to ensure bidirectional uniqueness. This is the approach in the reference Java solution. However, your solution is also correct and efficient.
3. Consider edge cases: what if the strings contain non-alphabet characters? Your solution should work for any ASCII character, which it does because the dictionary and set can handle any character.

Here is a slightly optimized version without the length check:

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        mp = {}
        seen = set()
        for i in range(len(s)):
            if s[i] not in mp:
                if t[i] in seen:
                    return False
                mp[s[i]] = t[i]
                seen.add(t[i])
            elif mp[s[i]] != t[i]:
                return False
        return True

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

VERDICT: PASS

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Word Pattern (word-pattern.py)

Your solution is correct and efficient. You have correctly implemented the bijection check using two dictionaries, which is the standard approach for this problem. The time and space complexity are optimal. The code is clear and well-commented, which is excellent.

However, there are a few minor points for improvement:

1. Variable names: Consider using more descriptive names. For example, instead of a, you could use char or pattern_char. Instead of arr, you could use words to make the code more readable.
2. The comment for sMap has an error in the example (it should map "cat" to 'b' in the example pattern "abba"), but since it's just a comment, it doesn't affect the code. Still, accurate comments are helpful.
3. You can avoid creating the list arr if you use an iterator, but since the problem requires splitting the string, your approach is straightforward and acceptable.

Overall, great job! Your solution is correct and follows best practices.

VERDICT: PASS