In a min-heap of n elements, implemented with a complete binary tree, the worst case time cost for removeMin() is

Question

BlackTom AI · Accepted Answer

Removing the minimum from a min-heap requires replacing the root with the last element and then restoring the heap order by repeatedly sifting that element down the tree. The height of a complete binary tree with n nodes is h = floor(log2 n), so the sift-down can traverse at most h levels.

1. Relevant concepts
- Min-heap: a complete binary tree where each node ≤ its children; implemented in an array.
- Height of complete binary tree: h = floor(log2 n).
- removeMin() steps: remove root, move last element to root, decrease size, then sift-down (heapify) to restore heap property by swapping with the smaller child until the node is ≤ both children or at a leaf.

2. Step-by-step reasoning
- Removing the root and moving the last element to the root are O(1) operations.
- The sift-down may compare and swap the current node with one child at each level. Each comparison/swap is O(1).
- In the worst case the element moves from the root to a leaf, which is at most h levels.

3. Intermediate calculations
- Height h = floor(log2 n). Example: if n = 15, h = floor(log2 15) = 3; worst-case sift-down does up to 3 swaps.
- Total worst-case work = O(1) (remove + replace) + O(h) (sift-down) = O(h) = O(log n).

4. Why the correct answer is correct
- Because the restoring operation is bounded by the tree height, and the height grows logarithmically with n, the worst-case time complexity is O(log n). This is a standard result for heap removeMin (also known as extract-min).

5. Why other options are incorrect
- O(1): incorrect because sift-down may require multiple swaps along the height, not constant time.
- O(n): incorrect because you never need to touch all n elements—only at most one element per level (∼log n levels).
- O(n^2): grossly incorrect; no quadratic behavior occurs for a single remove operation.

6. Conclusion
The worst-case time cost for removeMin() in a min-heap of n elements is O(log n), which matches the provided answer.

In a min-heap of n elements, implemented with a complete binary tree, the worst case time cost for removeMin() is单项选择题

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