BinarySearch2

[MeghaN28]

No description provided.

[super30admin]

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

Note: The verdict should be PASS if the solution is correct, efficient, and follows good practices. Otherwise, it should be NEEDS_IMPROVEMENT.

Now, evaluate the student's solution.

VERDICT: NEEDS_IMPROVEMENT

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Find the Minimum Element in a Rotated Array(sorted) (LowestElementRotatedArr.java)

Your solution is well-implemented and correctly addresses the problem. Here are some strengths and minor suggestions:

Strengths:

• You correctly handle the base case where the array is already sorted (nums[low] <= nums[high]).
• You properly check if the mid element is the minimum by comparing with neighbors, including boundary checks.
• You effectively use binary search to reduce the problem size by half each time, ensuring O(log n) performance.
• The code is readable and well-commented, which helps in understanding the logic.

Areas for Improvement:

• The condition for checking if mid is the minimum could be simplified. The reference solution uses the same condition, but note that when the array has only one element, mid==0 and mid==nums.length-1, so the condition holds. However, it's correct as written.
• Consider adding a fallback return statement at the end (like return -1) for completeness, even though the problem constraints ensure we always find a solution. Your code already has this, which is good.
• The comment "Confused about checking the opposite of sorted side" suggests you initially struggled with this. Remember: the minimum element always lies in the unsorted part. If the left side is sorted (nums[low] <= nums[mid]), then the right side must be unsorted and contain the minimum, so you move low to mid+1. Conversely, if the right side is sorted, the left side must contain the minimum, so you move high to mid-1. Your code correctly implements this.

Overall, excellent work! Your solution is efficient and correct.

VERDICT: PASS

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Find the Peak Element (FindPeak.java)

Your solution is well-implemented and efficient. It correctly uses binary search to achieve O(log n) time and O(1) space. The code is readable and you have provided good comments.

One minor point: In the else-if condition, you are accessing nums[mid+1] without explicitly checking if mid is the last element. However, your initial condition already ensures that when mid is the last element, it is handled as a peak and returns. So the code is safe. However, for clarity and to make it more robust (in case the first condition is modified), you could consider adding a check like mid < nums.length-1 in the else-if condition. But it is not strictly necessary here.

Another small note: The problem states that the array may contain multiple peaks and returning any one is fine. Your solution handles this correctly by moving towards the higher neighbor, which guarantees finding a peak.

Overall, great job! Your solution is correct and efficient.

VERDICT: PASS