Use a sliding window with two pointers to maintain a window that can be made strictly increasing by removing at most one element. Track the longest valid window and its indices, updating when a longer window is found or when the same length yields a lexicographically smaller start index. Ensure O(n) time by moving pointers only forward and O(1) space by using a few variables.
Pro tip: Clarify that 'removing at most one element' means the resulting subarray must be strictly increasing after removal, and that the indices refer to the original array. Emphasize that the lexicographically smallest pair is determined by comparing the start index first, then the end index.
Restate the problem: find the longest contiguous subarray that becomes strictly increasing after removing at most one element, and return its length and [l, r] in the original array. Note the O(n) time and O(1) space constraints.
Use two pointers (left and right) to represent the current window. Maintain a count of 'bad' adjacent pairs (where arr[i] >= arr[i+1]) within the window. Expand right and shrink left as needed to keep the count ≤ 1.
When the window is valid (bad count ≤ 1), compute its length. If it's longer than the current best, update best length and indices. If equal, compare start indices and update if the new start is smaller.
Consider arrays of length 0 or 1, and ensure that when multiple windows have the same length, the one with the smallest starting index is chosen. Also, verify that removing one element indeed makes the window strictly increasing.
Explain that each element is visited at most twice (by left and right pointers), giving O(n) time. Space is O(1) as only a few variables are used. Discuss potential pitfalls like off-by-one errors in index tracking.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.