Finished BinarySearch-2

[vishalmgvk]

No description provided.

[super30admin]

Find the First and Last Position of an Element in given Sorted Array (P1.py)

• Strengths: The binary search implementation for the peak element problem is well-structured with proper boundary checks and O(log n) time complexity.
• Weaknesses: The solution addresses the wrong problem entirely. The student needs to implement two separate binary searches - one for finding the first occurrence and one for finding the last occurrence of the target.
• Key improvements needed:
  1. Implement the correct algorithm for finding first and last positions
  2. Use two binary searches or modify the approach to track both boundaries
  3. Return a list/array with two values instead of a single index
  4. Add proper imports

VERDICT: NEEDS_IMPROVEMENT

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Find the Minimum Element in a Rotated Array(sorted) (P2.py)

Strengths:

• Correctly implements the binary search algorithm for finding minimum in rotated array
• Clean, readable code with appropriate comments
• Handles edge cases (checking mid > 0 before accessing nums[mid-1])
• Optimal time and space complexity

Areas for improvement:

• The condition on line 13 could be enhanced to explicitly check if mid is the minimum by also verifying mid == n-1 || nums[mid] < nums[mid + 1], similar to the reference solution. This makes the logic more comprehensive.
• Consider adding a docstring to the method explaining its purpose and parameters for better documentation.

Overall, this is a solid solution that correctly solves the problem with optimal complexity.

VERDICT: PASS

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Find the Peak Element (P3.py)

• Major Issue: You've solved the wrong problem entirely. The problem asks for finding a peak element (greater than neighbors), not searching for a target value's range.
• Strengths: Your code is well-structured, properly handles edge cases, and uses good binary search patterns for the problem you attempted.
• What to fix: You need to implement a binary search that compares the middle element with its neighbors to determine which direction to search. A peak element is greater than both its neighbors (or at boundaries).
• Key insight: For peak element, if nums[mid] < nums[mid+1], there must be a peak to the right; otherwise, search left.

VERDICT: NEEDS_IMPROVEMENT