Associate Software Engineer

Step 1:
Initially, I considered using linear search. That would take O(n) time, but since the problem required O(log n), this approach was not optimal.

Step 2:
Since the array was originally sorted, I realized that binary search could be used. However, due to rotation, the array is split into two sorted halves.

Step 3:
During binary search, I calculated mid = (low + high) / 2.
Then, I checked which half (left or right) was sorted:

If nums[low] ≤ nums[mid], then the left half is sorted.
Otherwise, the right half is sorted.

Step 4:
If the target lies within the sorted half, I moved the search to that half.
Otherwise, I searched in the other half.

Step 5:
I repeated this process until I found the target or the search space became empty.

This approach maintains O(log n) time complexity.

Step 1:
Initially, I considered checking all possible pairs of lines using two nested loops. This brute-force approach would take O(n²) time, which is inefficient for large inputs.

Step 2:
Then, I observed that the width is determined by the distance between two pointers, and the height is determined by the minimum of the two heights.

Area = min(height[left], height[right]) × (right - left)

Step 3:
I applied the two-pointer approach:

Initialize left = 0
Initialize right = n - 1

Step 4:
Calculate the area for the current pointers. Then move the pointer with the smaller height inward because:

• The area is limited by the smaller height.
• Moving the taller one will not help increase the area.

Step 5:
Repeat this process while left < right, and keep updating the maximum area.

This reduces the time complexity to O(n).

Step 1:
Initially, I considered storing visited nodes in a HashSet. While traversing the linked list, if a node was already present in the set, it indicates a cycle. This approach works but requires O(n) extra space.

Step 2:
Then, I optimized the solution using Floyd’s Cycle Detection Algorithm (the Tortoise and Hare approach), which works in O(1) space.

Step 3:
I used two pointers:

• The slow pointer moves one step at a time.
• The fast pointer moves two steps at a time.

Step 4:
If there is a cycle, the fast pointer will eventually meet the slow pointer.
If the fast pointer reaches null, then there is no cycle.

This optimized approach runs in O(n) time and O(1) space.

I answered honestly and kept my responses structured and concise. For strengths and weaknesses, I provided practical examples instead of generic answers. I focused on demonstrating a growth mindset, a willingness to learn, and consistency in my efforts. Most importantly, I maintained confidence, clarity, and a positive attitude throughout the conversation.