The examples made it feel approachable at first.
Use a monotonic stack to find the nearest greater element to the left and right for each index, then for each pair of indices that can be endpoints, check if the minimum in between is less than both endpoints. Optimize by considering only candidate pairs where the distance is maximized, and use a segment tree or sparse table for range minimum queries to achieve O(n log n) or O(n) time.
Pro tip: Clarify the problem constraints and edge cases (e.g., all equal elements, strictly increasing/decreasing arrays) before diving into the solution. Mention that while O(n log n) is acceptable, an O(n) solution using a monotonic stack and clever pruning exists, showing depth.
Restate the problem in your own words: find the longest subarray of length ≥2 where both endpoints are strictly greater than all elements between them. Confirm input size and expected complexity.
Note that the condition implies the endpoints are the two largest elements in the subarray. The subarray must have at least one element between, and all between must be smaller than both endpoints.
Use a monotonic stack to compute for each index the nearest greater element to the left and right. Then, for each index as a potential endpoint, consider the farthest index to the right that is greater and has all elements between smaller than both. Use range minimum queries to validate.
Explain that the monotonic stack takes O(n) time, and range minimum queries can be answered in O(1) with sparse table after O(n log n) preprocessing, leading to O(n log n) overall. Alternatively, a two-pointer or stack-based approach can achieve O(n).
Walk through examples like [5,1,4] (length 3), [1,2,3] (no valid subarray), and arrays with duplicates. Discuss how the algorithm handles them.
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