Completed Binary Search 2

[Keerthi0910]

No description provided.

[super30admin]

Find First and Last Position of Element in Sorted Array (Problem1.java)

EVERAGE_SCORE: [score from 1-10, where (log n) is best, O(n) is okay, O(n^2) or worse is bad]

VERDICT: PASS

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Find Minimum in Rotated Sorted Array (Problem2.java)

Strengths:

1. Your solution correctly implements binary search with O(log n) time complexity.
2. The early return optimization (nums[low] <= nums[high]) is a nice touch that handles the case when the array is already sorted.
3. Good code structure and readability.
4. You correctly identified the pivot point detection logic.

Areas for Improvement:

1. Boundary check consistency: Consider using mid == 0 instead of mid == low for the left boundary check. While mid == low works, mid == 0 is more semantically clear about checking the array's actual left boundary. This makes the intent clearer to readers.

2. Edge case consideration: You might want to add a brief comment explaining the logic, especially the conditions for identifying the minimum element. This would help future readers (and yourself) understand the algorithm.

3. Code comments: While you noted the issue you faced, adding inline comments explaining the key conditions (e.g., why we check nums[mid] < nums[mid-1]) would improve maintainability.

4. Minor style: Consider adding a space after if for consistency: if (condition) rather than if(condition).

Overall, this is a solid solution that demonstrates a good understanding of binary search in rotated sorted arrays.

VERDICT: PASS

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Find Peak Element (Problem3.java)

Your solution is correct and well-implemented! Here are some observations:

Strengths:

• You correctly translated the binary search approach from the reference solution
• Good edge case handling for empty arrays and single-element arrays
• The peak detection logic is properly implemented with boundary checks
• The code is readable and well-commented
• Time and space complexity match the optimal solution

Areas for improvement:

• The comment "If low is less than mid then low = mid+1" is slightly misleading - it should describe the actual condition (when nums[mid+1] > nums[mid])
• Consider adding more descriptive comments explaining why we move low up or high down (the intuition behind the binary search)
• The return -1 at the end is technically unreachable since a peak always exists in a non-empty array (given the problem constraints), but it's good defensive programming

Minor suggestions:

• You could simplify the edge case handling by noting that the problem guarantees at least one peak exists
• The code could benefit from a brief explanation of the algorithm's correctness in comments

Overall, this is a solid solution that demonstrates good understanding of the binary search technique for finding peak elements.

VERDICT: PASS