Turn the following array into a binary max-heap using the heapify operation.

Question

BlackTom AI · Accepted Answer

The provided data for this question is incomplete.
- There is no array given to convert into a binary max-heap, so I cannot perform the actual heapify steps on a concrete sequence.
- There are no answer options listed either, so I cannot analyze which choice would be correct or incorrect.

What I can do is outline the standard approach to turning a given array into a binary max-heap using heapify (sift-down) operations, so you have a reference when you do have the actual array.

General procedure for heapify into a max-heap (bottom-up approach):
- Let the array be A[0..n-1]. A max-heap requires that for every node i, A[i] >= A[left(i)] and A[i] >= A[right(i)], where left(i) = 2*i + 1 and right(i) = 2*i + 2.
- Start from the last internal node, which is at index floor(n/2) - 1, and move leftward to index 0. For each index i, perform a sift-down (heapify-down) operation:
  - Find the largest among A[i], A[left(i)], and A[right(i)].
  - If the largest is not A[i], swap A[i] with A[largest], and recursively apply heapify-down to the index where the swap occurred.
- After processing all internal nodes this way, the array will satisfy the max-heap property.

Why this bottom-up method works:
- Leaves are already heaps of size 1, so you only need to fix internal nodes.
- Each sift-down operation guarantees that the subtree rooted at the current index becomes a valid max-heap, and since you process from bottom up, previously fixed subtrees remain valid.

Common details to keep in mind:
- If the left and right children are out of bounds, you’re at a leaf and can stop.
- When both children exist, compare both children to determine the larger child, then compare with the parent to decide whether to swap.
- If there is a tie for the larger value between children, some implementations choose the left child as the larger by convention, but the exact choice does not affect the correctness of the resulting heap as long as you consistently pick the larger one during each step.

If you can share the actual array (and any specific answer choices if this is a multiple-choice item), I can walk through the exact heapify-down steps for each relevant index and explain why the final arrangement forms a valid max-heap, addressing each option in turn.

Turn the following array into a binary max-heap using the heapify operation.

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