My first instinct was two pointers and I started coding before really thinking through what happens when the drop occurs mid-window.
Use a sliding window with two pointers to maintain a window that is strictly increasing except for at most one drop. Expand the right pointer, and when a second drop is encountered, move the left pointer to just after the first drop to restore validity. Track the maximum window length throughout.
Pro tip: Clarify edge cases upfront, such as empty arrays or arrays with all equal elements, and mention that the solution runs in O(n) time and O(1) space, which is optimal for this problem.
Restate the problem: find the longest contiguous subarray that is strictly increasing except for at most one drop. Ask clarifying questions about edge cases and input size.
Select a sliding window approach because it efficiently tracks a valid window and adjusts when the condition is violated. Mention that a brute-force check would be O(n^2) and is not optimal.
Maintain a window [left, right] that is valid. When adding arr[right], if arr[right] <= arr[right-1], increment a drop counter. If drop counter exceeds 1, move left to the position after the previous drop (i.e., left = last_drop_index + 1) and reset the drop counter appropriately.
Iterate right from 0 to n-1, update the window, and after each step update max_len = max(max_len, right - left + 1). Handle the case when a drop occurs by updating last_drop_index.
State that the algorithm runs in O(n) time and O(1) space. Walk through a few test cases, including arrays with no drops, one drop, multiple drops, and edge cases like empty or single-element arrays.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.