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Find First and Last Position of Element in Sorted Array (findPeakElement.java)
The code correctly implements binary search, but for a completely different problem
The binary search implementation itself is clean and follows best practices (using low + (high-low)/2 to avoid overflow)
Good variable naming and code structure
However, the student appears to have submitted code for the wrong LeetCode problem
To solve the actual problem, the student needs to implement two binary searches: one for the first occurrence and one for the last occurrence of the target
VERDICT: NEEDS_IMPROVEMENT
Find Minimum in Rotated Sorted Array (findMin.java)
Strengths:
Correct binary search logic for finding minimum in rotated array
Efficient O(log n) time and O(1) space complexity
Clean and concise implementation
Proper handling of all test cases
Areas for Improvement:
Follow Java naming conventions: class names should start with uppercase (Solution)
Add comments to explain the logic, especially the key insight about comparing nums[mid] with nums[right]
Consider adding a check for edge cases like empty arrays if the problem constraints allowed them
VERDICT: PASS
Find Peak Element (searchRange.java)
Read the problem carefully: You submitted a solution for a completely different problem. The "Find Peak Element" problem requires finding an element greater than its neighbors, not searching for a target value.
Fix syntax errors: "pblic" should be "public" - this is a basic typo that would cause compilation failure.
Fix method signatures: The problem asks for a single index return, but your solution returns an int array.
Fix logic errors: In findLast, the condition else if(nums[mid]<target) is redundant and incorrect. Also, methods don't return values in all paths.
Study the correct approach: For peak element finding, use binary search comparing nums[mid] with nums[mid+1] to determine which direction to search.
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