Completed Binary-Search-2

[samikshm]

No description provided.

[super30admin]

Find the First and Last Position of an Element in given Sorted Array (find-first-and-last-position-of-element-in-sorted-array.java)

Strengths:

1. The solution correctly implements two binary search functions to find the first and last occurrences, which is efficient and meets the O(log n) runtime requirement.
2. The code is well-commented and follows good practices, such as using private helper methods and meaningful variable names.
3. The solution handles edge cases, such as when the target is not found, by returning [-1, -1].

Areas for Improvement:

1. The binarySearchLast function can be optimized by starting the search from the first occurrence index (as done) to the end, which is efficient. However, note that in the worst case, if the entire array is the target, the second binary search will still be O(log n), which is acceptable.
2. There is a minor improvement: in binarySearchLast, the initial low parameter is set to the first occurrence index, which is correct and avoids unnecessary searches in the left part. This is a good optimization.
3. The code is clear, but you might consider adding a brief comment in the main method to explain the flow, although the existing comments are sufficient.

Overall, the solution is correct and efficient.

VERDICT: PASS

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Find the Minimum Element in a Rotated Array(sorted) (find-minimum-in-rotated-sorted-array.java)

It seems that you haven't submitted any code for this problem. To solve this problem, you need to implement a binary search algorithm that runs in O(log n) time. The key idea is to compare the middle element with the left and right ends to determine which half of the array contains the minimum element. Here are some steps to guide you:

1. Initialize two pointers, low at the start and high at the end of the array.
2. While low <= high:
   • If the element at low is less than or equal to the element at high, it means the current segment is sorted, so the minimum is at low.
   • Calculate the mid index.
   • Check if the mid element is the minimum by comparing it with its neighbors (if they exist). If it is smaller than both neighbors, return it.
   • Otherwise, if the left half (from low to mid) is sorted, then the minimum must be in the right half, so set low to mid+1.
   • Else, the right half is sorted, so the minimum must be in the left half, so set high to mid-1.

Please try to implement the solution and resubmit. If you need help with the code, you can refer to the reference solution provided.

VERDICT: NEEDS_IMPROVEMENT

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Find the Peak Element (find-peak-element.java)

Your solution is excellent and follows the same logic as the reference solution. You correctly check if the current mid is a peak by comparing it with its neighbors, and then move towards the increasing slope to find a peak. The code is clean and efficient. One minor point: in the condition for the increasing slope, you assume that mid+1 exists because you are within the while loop and have already checked boundaries, but it's good practice to note that the condition mid < nums.length - 1 is implicitly handled by the initial peak check. However, to avoid any potential issues (though none exist here), you could consider adding a boundary check in the else-if condition, but it is not necessary since the peak condition already ensures that if mid is not the last element, we can check mid+1. Overall, great job!

VERDICT: PASS