Completed leetcode 702, 74 and 33

[allurkarsneha]

No description provided.

[super30admin]

Search in Rotated Sorted Array (leetcode33.py)

[Portfolio feedback for the student]

EPORTFOLIO_FEEDB

VERDICT: NEEDS_IMPROVEMENT

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Search in a Sorted Array of Unknown Size (leetcode702.py)

EULTIMATE
EVALUATION:
The student's solution correctly implements the algorithm for searching in a sorted array of unknown size. Let me analyze it:

1. Correctness: The solution is correct. It uses the same approach as the reference solution:

   • First, it exponentially expands the search range by doubling h until reader.get(h) >= target
   • Then, it performs binary search within the established range
   • The logic handles all cases properly, including when the target is not found

2. Time Complexity: O(log n) - matches the reference solution. The exponential expansion phase takes O(log n) time, and the binary search phase also takes O(log n) time.

3. Space Complexity: O(1) - matches the reference solution. Only constant extra space is used.

4. Code Quality:

   • The code is well-structured and readable
   • Variable names are clear (l, h, mid)
   • The docstring provides good documentation
   • The comments at the top indicate time/space complexity analysis
   • Python conventions are followed

5. Efficiency: The solution is efficient. One minor consideration is that reader.get(h) is called multiple times in the loop, but this is necessary for the algorithm. The solution is essentially identical to the reference solution in terms of approach.

The student's solution is essentially a direct translation of the reference solution from C++ to Python, which is perfectly acceptable. The algorithm is correct and efficient.

FEEDOK:
Your solution is excellent! It correctly implements the two-phase approach:

1. Exponential search to find the upper bound
2. Binary search within the established range

Strengths:

• Clean, readable code with appropriate variable names
• Good documentation with docstring and complexity analysis
• Correct handling of edge cases
• O(log n) time complexity and O(1) space complexity as required

Minor suggestions for improvement:

• Consider adding a brief comment explaining the two-phase approach for educational purposes
• The variable name h could be more descriptive (e.g., high or upper_bound) for better readability, though this is a minor stylistic preference

Overall, this is a solid solution that demonstrates a clear understanding of the problem and the appropriate algorithm.

VERDICT: PASS

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Search a 2D Matrix (leetcode74.py)

EPORT_TEMPLATE:

# Evaluation Report

## Problem
**Title:** Search a 2D Matrix  
**Difficulty:** Medium

## Reference Solution
- **Time Complexity:** O(log m*n)
O(1)
- **Space Complexity:** O(1)
- **Language:** Java

Reference solution uses binary search treating the matrix as a virtual 1D array of size m*n. It computes row and column indices from the mid index using integer division and modulo operations.

## Student Solution
- **Time Complexity:** O(log(m * n))
- **Time Complexity:** O(1)
- **Space Complexity:** O(1)
- **Language:** Python

The student solution uses the same approach as the reference solution - treating the matrix as a virtual 1D array and applying binary search. The implementation is correct and efficient.

## Analysis

### Correctness
The student's solution is correct. It properly:
- Initializes search bounds `l = 0` and `h = m * n - 1`
- Computes mid index correctly
- Maps mid index to row and column using `mid // n` and `mid % n`
- Updates search bounds correctly based on comparison with target
- Returns `True`/`False` appropriately

### Time Complexity
O(log(m * n)) - matches the reference solution.

### Space Complexity
O(1) - matches the reference solution.

### Code Quality
The code is well-structured with:
- Clear docstring describing parameters and return type
- Appropriate variable names
- Proper indentation
- Comments indicating complexity analysis

### Efficiency
The solution is optimal for this problem.

## Verdict
**PASS**

Now, please provide your detailed evaluation and feedback.

VERDICT: NEEDS_IMPROVEMENT