Classic heap problem once you see it, but I stared at it for a bit before the min-heap idea clicked.
Recognize that the array is nearly sorted with each element at most K positions from its correct position. Use a min-heap of size K+1 to efficiently extract the smallest element and place it in the correct position, achieving O(N log K) time. Alternatively, use insertion sort which is O(NK) but may be simpler; however, the heap approach is optimal for large N and small K.
Pro tip: Mention that for K=1 (almost sorted), insertion sort is O(N) and simpler, but for general K, the heap approach is better. Also, note that if K is large (close to N), just use a standard O(N log N) sort.
Clarify that each element is at most K positions away from its sorted position. Discuss the implications: the array is 'nearly sorted', and we can exploit this for efficiency.
Decide between insertion sort (O(NK)) and min-heap (O(N log K)). For large N and small K, heap is better. Explain why heap works: the smallest element among the first K+1 elements must be the global minimum.
Initialize a min-heap with the first K+1 elements. Then, for each subsequent element, extract the minimum from the heap and place it in the array, then add the new element to the heap. Finally, extract remaining elements.
Time complexity: O(N log K). Space: O(K). Handle edge cases: K=0 (already sorted), K>=N (just sort normally), and ensure heap size does not exceed array bounds.
Walk through a small example (e.g., arr = [2,1,4,3], K=1) to verify correctness. Discuss potential pitfalls like off-by-one errors in heap size.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.