I kept the items in a sorted list using bisect insertion, so each add is O(log n).
Clarify the problem requirements, then propose an efficient data structure like a balanced BST or sorted list to maintain items in sorted order. For each new item, check windows of three consecutive items in the sorted order to see if the difference between max and min is below the threshold, returning the first such triple.
Pro tip: Mention that checking only consecutive triples in sorted order is sufficient because any valid triple must have its max and min within a window of three consecutive elements. This shows deep insight and avoids unnecessary complexity.
Ask about input constraints, threshold definition, and whether the triple should be returned immediately or if all triples need to be found. Confirm if items are ingested one at a time and if duplicates are allowed.
Select a data structure that maintains sorted order and supports efficient insertion and neighbor queries, such as a balanced binary search tree (e.g., TreeSet in Java) or a skip list.
On each add(size), insert the new size into the sorted structure. Then check the new item's immediate neighbors (up to two on each side) to form all possible consecutive triples involving the new item. For each triple, compute max - min and compare to threshold.
If a valid triple is found, return it immediately. Discuss time complexity: O(log n) per insertion and O(1) checks per insertion, leading to O(n log n) overall. Mention potential optimizations like early termination.
Address cases with fewer than three items, duplicate sizes, and threshold boundaries. Ensure the solution works for large streams and discuss memory considerations.
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