I knew immediately this was a two-heap setup, but the part that tripped me up was maintaining balance while sliding the window.
Use a balanced binary search tree (or two heaps) to maintain the sliding window of size k, supporting O(log k) insertions and deletions. For each window, compute the lower median in O(1) or O(log k) time, then track the maximum and minimum of these medians. This achieves O(n log k) time and O(k) space.
Pro tip: Clarify the lower median definition upfront and mention that using a single balanced BST with order statistics (e.g., an order-statistic tree) simplifies median retrieval to O(log k). Also, note that two heaps require lazy deletion, which adds complexity but is still O(log k) amortized.
Confirm the definition of lower median for even k (e.g., the smaller of the two middle elements). Discuss edge cases: k=1, k=n, and when n<k (though typically n>=k).
Select a balanced BST with order statistics (e.g., an order-statistic tree) or two heaps (max-heap for lower half, min-heap for upper half) to maintain the sliding window. Explain why this gives O(log k) per operation.
Insert the first k elements into the structure, compute the median, and update max/min. Then for each subsequent element, remove the outgoing element and insert the new one, recomputing the median each time.
After each window update, retrieve the lower median (e.g., the (k+1)/2-th smallest for odd k, or k/2-th smallest for even k). Compare it to the current max and min, updating as needed.
Summarize that each insertion/deletion takes O(log k), and there are O(n) windows, giving O(n log k) time. Space is O(k) for the data structure. Mention alternative approaches and why they don't meet the constraints.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one felt like a DP problem dressed up in a weird costume.
First, recognize that each segment must be a pair of identical bits (since even length and all identical values implies length 2). Thus, the problem reduces to partitioning the array into adjacent pairs and flipping bits within each pair to make them equal. To minimize segments, we want as few pairs as possible, which means maximizing the length of each segment—but since segments must be even and uniform, the only possible segment length is 2. So the minimum number of segments is n/2, and the minimum flips is the sum over pairs of the cost to make each pair uniform (0 if already equal, 1 otherwise). The algorithm is simply to iterate through the array in steps of 2, count mismatches, and return n/2 segments and the mismatch count as flips. Complexity is O(n) time and O(1) space.
Pro tip: Clarify early that 'even length and all identical values' forces each segment to be exactly length 2, which simplifies the problem drastically. This shows you can cut through ambiguity and focus on the core constraint.
Confirm that segments must be contiguous, even-length, and contain identical values. Ask if the array can be partitioned arbitrarily or if the partition must cover the entire array.
Reason that an even-length segment with all identical bits must have length 2, because any longer even length would require more than two identical bits, which is impossible in a binary array without violating the 'all identical' condition for the whole segment.
Since each segment is length 2, the number of segments is fixed at n/2. Minimizing segments is trivial; then minimize flips by making each pair uniform with minimal changes.
Iterate through the array in steps of 2. For each pair, if the bits differ, increment a flip counter. Return n/2 as the segment count and the flip counter as the minimum flips.
State that the algorithm runs in O(n) time and O(1) extra space, as it only requires a single pass and a constant number of variables.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.