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Spent way too long on the naive approach, checking each window independently which is obviously O(nk).
First, clarify the problem: for each window of size k (odd), check if the middle element is strictly greater than its immediate neighbors, and the sequence strictly decreases as you move away from the center in both directions. Then, propose an O(n) solution using precomputed arrays that track the length of decreasing runs to the left and right of each index, and for each window center, check if the required decreasing lengths are met. Finally, analyze the time and space complexity, emphasizing the O(n) time and O(n) space trade-off.
Pro tip: Mention that this is essentially a 'mountain array' check for each window, and that by precomputing the longest decreasing run ending at each index and the longest decreasing run starting at each index, you can answer each window in O(1) time, achieving overall O(n). Also, note that the problem can be solved in O(n) time with O(n) extra space, or O(n) time with O(1) extra space if you use a sliding window with two pointers, but the precomputation method is simpler to implement and explain.
Confirm that the window size is odd, and that 'bitonic centered on the middle' means strictly decreasing from the center to both ends. Discuss edge cases: window size 1 (always true), array length less than window size, and handling of equal adjacent elements (should break the decreasing condition).
For each window, check the bitonic property by scanning from the center outwards. This takes O(n*k) time, which is O(n^2) if k is O(n). Acknowledge that this is not optimal.
Create two arrays: left[i] = length of the longest strictly decreasing contiguous subarray ending at i (including i), and right[i] = length of the longest strictly decreasing contiguous subarray starting at i (including i). For a window centered at i with half-size h = (k-1)/2, the window is bitonic if left[i] >= h+1 and right[i] >= h+1. Iterate over all possible centers and collect positions where this holds.
Precomputation takes O(n) time and O(n) space. Checking each center takes O(1) time, so total O(n) time. Space can be reduced to O(1) if we only need to check a fixed window size, but the precomputation method is straightforward.
Mention that if the window size is fixed, we can use a sliding window with two pointers to check the bitonic property in O(n) time and O(1) space, but it requires careful handling of the decreasing conditions. Also, note that the precomputation method can be adapted to return all valid centers efficiently.
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