Completed Binary Search-2 Assignment

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Find the First and Last Position of an Element in given Sorted Array (34_find_first_and_last_element.java)

Your solution is excellent! You have correctly implemented two binary search functions to find the first and last occurrences of the target. The code is efficient with O(log n) time complexity and O(1) space complexity.

Strengths:

• You used binary search effectively, ensuring the required time complexity.
• The code is clean and well-commented, making it easy to understand.
• You handled edge cases, such as when the target is not found (returning [-1,-1]) and when the array has only one occurrence of the target.

Areas for improvement:

• While the code is correct, you can consider adding more comments to explain the binary search conditions for clarity, especially for someone learning the algorithm.
• Ensure consistent formatting (e.g., spaces around operators) to enhance readability. For example, in the condition "else if (nums[mid]>target)", adding a space like "nums[mid] > target" is more conventional.
• Although not necessary, you could make the binarySearchLast function start from the entire array range instead of from the first occurrence. However, starting from the first occurrence is an optimization that reduces the search range for the last occurrence, which is good.

Overall, this is a solid implementation. Keep up the good work!

VERDICT: PASS

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Find the Minimum Element in a Rotated Array(sorted) (154_rotated_sorted_array_min.java)

Strengths:

• The solution correctly implements binary search to achieve O(log n) time complexity.
• The code is well-structured and includes comments that show the student's thought process.
• The condition to check if the current segment is sorted (if nums[low] <= nums[high]) is correctly placed and efficient.
• The condition to check if the mid element is the minimum by comparing with neighbors is correctly implemented.

Areas for Improvement:

• The condition if (nums[mid] >= nums[low]) is used to determine if the left half is sorted. However, note that when the array has only two elements, mid might be equal to low, and this condition holds. This is correct, but it's important to note that the condition should be nums[low] <= nums[mid] to be consistent with the reference solution. The student's condition nums[mid] >= nums[low] is equivalent and correct.
• The code could benefit from a brief comment explaining the main logic steps, which would enhance readability.
• The fallback return of -1 is appropriate, but it's worth noting that the problem constraints guarantee a solution, so this is safe.

Overall, the solution is robust and handles all edge cases, including when the array is not rotated (fully sorted) and when the minimum is at the beginning or end.

VERDICT: PASS

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Find the Peak Element (162_find_peak_element.java)

Strengths:

• The solution correctly implements the binary search algorithm for finding a peak element.
• The time and space complexity are optimal.
• The code is clean and easy to follow.

Areas for Improvement:

• The condition for checking the peak element is correct, but it could be made more readable by breaking it down or adding comments to explain the two parts (left neighbor and right neighbor). However, given the simplicity, it is acceptable.
• The solution assumes that the array has at least one element. Since the problem states that the array has elements and the constraints are such that we can assume valid input, this is fine. But it's always good to consider edge cases explicitly, though the code handles them with the conditions mid==0 and mid==nums.length-1.
• The code returns -1 if no peak is found, but according to the problem, there is always at least one peak. So this is safe, but the return -1 is never reached in practice. You might consider whether it's necessary or if you could assert that a peak always exists.

Overall, the solution is excellent and meets all requirements.

VERDICT: PASS