The k constraint is basically screaming 'use a heap' but I spent an embarrassing amount of time second-guessing myself before committing to that.
Use a min-heap of size k+1 to maintain the smallest candidates for the current position, then extract and insert the next element. This yields O(n log k) time and O(k) space, which is more efficient than O(n log n) when k is small.
Pro tip: Emphasize that the heap size is k+1, not k, because each element can be at most k positions away, so the correct element for the current position is within the next k+1 elements. Also, mention that if k is large (e.g., k = n), the heap approach degrades to O(n log n), so it's important to consider the trade-off.
Restate that each element is at most k positions from its sorted position, meaning the array is 'k-sorted'. Clarify that k is typically much smaller than n, and the goal is to achieve better than O(n log n).
Select a min-heap to efficiently retrieve the minimum among the next k+1 elements. Explain why a heap is suitable: it provides O(log k) insertion and extraction.
Initialize a min-heap with the first k+1 elements. Then, for each position from 0 to n-1, extract the minimum and place it in the sorted array, and if there are remaining elements, insert the next element into the heap.
Time: O(n log k) because each of the n elements is inserted and extracted once, each operation O(log k). Space: O(k) for the heap. Compare with O(n log n) and note that for k << n, this is a significant improvement.
Mention edge cases: k=0 (already sorted), k=n (no improvement), and k close to n. Also, note that if k is large, other algorithms like quicksort might be preferable, and that the heap approach is stable if implemented carefully.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.