Completed Hashing-2

[satish-paraddi]

No description provided.

[super30admin]

Subarray Sum Equals K (Problem-1.py)

[Constructive feedback for you, highlighting strengths and areas for improvement]

VERDICT: PASS

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Contiguous Array (Problem-2.py)

EVIDENCE: [Reference solution logic vs student's solution logic - explain how the student's solution matches or differs from the reference solution]

EVIDENCE:
The student's solution uses the same prefix sum approach as the reference solution:

• Both initialize a hashmap with {0: -1} to handle the case when the entire array has equal 0s and 1s.
• Both iterate through the array, treating 0 as -1 and 1 as +1.
• Both check if the current prefix sum exists in the hashmap; if yes, calculate the length of the subarray; if no, store the current index.
• Both return the maximum length found.

The logic is identical to the reference solution. The only differences are syntactic (Python vs Java) and minor stylistic choices (using max as a variable name, which shadows the built-in max function in Python).

EVALUATION:

1. Correctness: The solution correctly solves the problem. The prefix sum approach with hashmap is correctly implemented. The logic of treating 0 as -1 and 1 as +1, and tracking the earliest occurrence of each prefix sum, is sound.

2. Time Complexity: O(n) - matches the reference solution. The algorithm makes a single pass through the array with O(1) hashmap operations.

3. Space Complexity: O(n) - matches the reference solution. In the worst case, the hashmap stores n+1 entries.

4. Code Quality:

   • The code is well-structured and readable.
   • The comments at the top are helpful.
   • Minor issue: Using max as a variable name shadows Python's built-in max() function. This is a bad practice and could lead to bugs if max() is needed later in the function.
   • The two if statements for nums[i] == 1 and nums[i] == 0 could be combined into an if-else for slightly better readability.
   • The if prefix_sum in d and if prefix_sum not in d could be simplified using setdefault() or a single check.

5. Eikiency: The solution is efficient. No major optimizations needed.

FEEDEACK:
Great job! Your solution correctly implements the prefix sum approach with a hashmap, matching the reference solution in both time and space complexity. Here are some suggestions for improvement:

1. Avoid shadowing built-ins: Don't use max as a variable name since it shadows Python's built-in max() function. Use max_len or result instead.

2. Simplify the logic: The two separate if statements for checking nums[i] == 1 and nums[i] == 0 can be combined into an if-else:

   if nums[i] == 1:
       prefix_sum += 1
   else:
       prefix_sum -= 1

3. Use setdefault() or get(): The check for whether prefix_sum is in the dictionary can be simplified:

   if prefix_sum in d:
       max_len = max(max_len, i - d[prefix_sum])
   else:
       d[prefix_sum] = i

4. Add type hints and docstring: Consider adding a docstring to explain the approach.

Overall, this is a solid solution that demonstrates a good understanding of the prefix sum technique.

VERDICT: PASS

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Longest Palindrome (Problem-3.py)

The student's solution correctly solves the problem. Let me analyze it:

1. Correctness: The solution is correct. It uses a set to track characters with odd counts. When a character is seen twice, it adds 2 to the length and removes it from the set. At the end, if there are any characters left in the set (odd counts), it adds 1 to allow one character to be placed in the middle of the palindrome.

2. Time Complexity: O(n) - same as the reference solution. Each character is processed once, and set operations are O(1) on average.

3. Space Complexity: O(1) - same as the reference solution. The set can contain at most 52 characters (26 lowercase + 26 uppercase), which is constant.

4. Code Quality: The code is clean, readable, and follows Python best practices. The variable names are descriptive, and the logic is clear. The comments at the top are helpful.

5. Efficiency: The solution is efficient and matches the reference solution's approach.

The student's solution is essentially identical to the reference solution, just translated from Java to Python. It correctly handles all edge cases.

FEEDirc:
Your solution is excellent! It correctly solves the problem with the same approach as the reference solution. Here are some observations:

Strengths:

• Clean and readable code
• Correct use of a set to track characters with odd counts
• Proper handling of the case where odd-count characters can be placed in the middle
• Good comments documenting time/space complexity
• The conditional expression return length + 1 if len(d) > 0 else length is a nice Pythonic touch

Minor suggestions:

• Consider adding a brief comment explaining the logic (e.g., "Use set to track chars with odd counts; pairs add 2, leftover odd chars can go in middle")
• The variable name d could be more descriptive (e.g., odd_chars or unpaired_chars)

Overall, this is a solid solution that demonstrates a good understanding of the problem.

VERDICT: PASS