add binary-search-2 solutions

[nikhylw]

No description provided.

[super30admin]

Find the First and Last Position of an Element in given Sorted Array (solution1.py)

Strengths:

• Clean implementation of binary search for both first and last positions
• Proper handling of edge cases (empty array, boundaries)
• Good use of nested helper functions for modularity
• Clear and readable code structure
• Correctly uses mid == 0 and mid == len(nums) - 1 checks to avoid index out of bounds

Areas for Improvement:

• The solution is functionally complete and matches the reference solution well
• Consider adding docstrings to the helper functions for better documentation
• Could potentially use Python's bisect_left and bisect_right for a more concise solution, though the current approach demonstrates stronger understanding of the underlying algorithm

VERDICT: PASS

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Find the Minimum Element in a Rotated Array(sorted) (solution2.py)

Strengths:

• The solution is elegant and more concise than the reference solution
• Correctly handles edge cases including arrays with no rotation
• Uses proper binary search with low + (high - low) // 2 to avoid potential overflow
• The logic of comparing with nums[high] instead of nums[low] is a common optimization that simplifies the implementation

Areas for Improvement:

• Consider adding a brief comment explaining the algorithm logic for future maintainability
• The solution is already optimal; no significant optimizations needed

Minor Note:
The condition nums[mid] <= nums[high] uses <= which correctly handles duplicate-like scenarios (though the problem states unique elements, this is a safe approach).

VERDICT: PASS

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Find the Peak Element (solution3.py)

Strengths:

1. The solution is concise and elegant - fewer lines of code than the reference
2. Correctly implements binary search for peak element
3. No explicit boundary checks needed due to smart loop condition
4. Clean variable naming and proper Python style

Areas for Improvement:

1. The solution could benefit from a brief comment explaining the binary search logic
2. Consider adding a comment about why mid + 1 access is safe (because low < high guarantees mid < high)

The solution is correct and efficient. It's actually slightly more elegant than the reference solution as it avoids explicit boundary checks by leveraging the loop condition.

VERDICT: PASS