Completed Hashing1

[SinAIML]

No description provided.

[super30admin]

Grouping Anagrams Together (group-anagrams.py)

Your solution is correct and well-explained. You have accurately analyzed the time and space complexity. However, there are a few points to improve:

1. Remove the print statement for temp as it is not needed and could lead to performance issues in production.
2. Consider optimizing the solution by using a more efficient hashing method, such as the prime number product approach, to avoid the O(k log k) sorting step. This would improve the time complexity to O(n * k).
3. The variable temp is assigned but not used; you can remove it to avoid confusion.
4. The solution is efficient for most cases, but for very large strings, the sorting might become a bottleneck. The prime number approach might be better in such cases.

Overall, your solution is good and passes all test cases, but with the noted improvements, it could be more efficient.

VERDICT: PASS

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Isomorphic Strings (isomorphic-strings.py)

Your solution has the right idea by using two dictionaries to track the mappings in both directions. However, there are several issues that need to be fixed:

1. Indentation: In Python, the code inside a function must be indented. Your entire function body is not indented, which will cause a syntax error. Please indent all lines inside the function.
2. Unnecessary Check: The line if len(smap) != len(tmap): is comparing the lengths of the dictionaries, which is not needed. You should remove this check.
3. Function Name: Consider using a more descriptive function name, such as isIsomorphic, to follow common naming conventions.
4. Edge Cases: Although the problem states that the strings have the same length, it's good practice to add a check for that. However, it is not strictly necessary.

Here is a corrected version of your code:

def isIsomorphic(s, t):
    if len(s) != len(t):
        return False
    smap = {}
    tmap = {}
    for i in range(len(s)):
        schar = s[i]
        tchar = t[i]
        if schar in smap:
            if smap[schar] != tchar:
                return False
        else:
            smap[schar] = tchar
        if tchar in tmap:
            if tmap[tchar] != schar:
                return False
        else:
            tmap[tchar] = schar
    return True

Your approach is correct in logic, but the indentation error would prevent it from running. Also, the initial check was misguided. Keep up the good work in thinking about bidirectional mapping!

VERDICT: NEEDS_IMPROVEMENT

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Word Pattern (word-pattern.py)

Your solution is well-structured and correctly solves the problem. You have correctly implemented the bijection check using a dictionary for pattern->word mapping and a set to ensure words are not reused. The code is readable with clear variable names.

Strengths:

• Correct handling of the bijection requirements.
• Efficient time and space complexity.
• Clear code structure and comments.

Areas for improvement:

• Remove the print statement for slist as it is unnecessary for the final solution and might be considered debugging code.
• Consider adding a check at the beginning to ensure the number of words in s matches the pattern length. You already have this check, which is good.
• You might want to consider using two dictionaries (one for pattern->word and one for word->pattern) to make the symmetry explicit. However, your current solution with a set is efficient and correct.

Overall, your solution is correct and efficient.

VERDICT: PASS