Use a single pass with two pointers or running counters to track the current increasing run and the best run seen so far. At each step, compare the current element with the previous one: if it's greater, extend the current run; otherwise, reset the current run to start at the current element. Update the best run whenever the current run exceeds it.
Pro tip: Clarify upfront that 'contiguous subarray' means a contiguous segment of the array, and confirm whether you need to return the subarray itself or just its length—this shows attention to detail and avoids wasted effort.
Confirm the definition of 'strictly increasing' and whether the output should be the subarray or its length. Discuss edge cases like empty array, single element, and all decreasing elements.
Explain that you'll iterate through the array once, maintaining the start index and length of the current increasing run, and the best run found so far.
For each element from index 1 to n-1, compare with the previous element. If it's greater, increment the current run length; otherwise, reset the current run to start at the current index with length 1. Update the best run if the current run is longer.
Initialize current run length to 1 and best run to 1 (or handle empty array separately). After the loop, return the best run (or the subarray using the recorded start and length).
State that time complexity is O(n) and space complexity is O(1) (or O(n) if returning the subarray). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem and confirm that the replacement can be any value, including values outside the current range. Then, propose an O(n) solution that precomputes the lengths of increasing runs ending and starting at each index, and for each potential replacement position, combines the run before and after if the gap can be bridged by a single value. Finally, discuss edge cases and trade-offs.
Pro tip: Mention that the replacement value can be chosen to satisfy the increasing condition if the difference between the boundary elements is at least 2, or if the replacement is at the array boundary. This shows attention to detail and avoids off-by-one errors.
Confirm that you can replace at most one element with any value, and that the subarray must be contiguous and strictly increasing. Ask if the replacement can be outside the original array's range.
Compute two arrays: inc_end[i] = length of longest strictly increasing subarray ending at i, and inc_start[i] = length of longest strictly increasing subarray starting at i. This can be done in O(n) time.
For each index i, consider replacing arr[i]. The best subarray including i is inc_end[i-1] + 1 + inc_start[i+1] if the gap between arr[i-1] and arr[i+1] can be bridged by a single value (i.e., arr[i+1] - arr[i-1] >= 2). Also consider replacing at boundaries (i=0 or i=n-1) where only one side exists.
If no replacement is used, the answer is the maximum of inc_end[i]. Also consider arrays of length 1 or 2. Return the maximum over all valid combinations.
The algorithm runs in O(n) time and O(n) space. Discuss if space can be reduced to O(1) by computing runs on the fly, but note that precomputation simplifies the logic.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.