The statement is **True**.
Let's break down the explanation in a systematic way:
1. **Converting Binary Min-Heap to a Sorted Array (O(n) comparisons):**
- In a binary min-heap, each node is smaller than its children, ensuring the minimum element is at the root.
- Extracting the minimum element and placing it in an array is a linear operation with O(n) comparisons. This is because, in each step, you compare and swap the root with the last element, then restore the heap property with a logarithmic number of comparisons (heapify).
2. **Building AVL Tree from the Sorted Array (O(n) comparisons):**
- Constructing an AVL tree from a sorted array involves selecting the middle element as the root, creating left and right subtrees recursively.
- Since the array is sorted, selecting the middle element can be done in constant time. The construction of the left and right subtrees involves traversing the remaining elements, which can be done in a linear fashion.
- The AVL tree construction maintains the balance property, ensuring the tree remains balanced after each insertion. The time complexity of constructing an AVL tree from a sorted array is O(n).
3. **Overall Time Complexity:**
- Combining the two operations, first converting the binary min-heap to a sorted array (O(n) comparisons) and then building an AVL tree from the sorted array (O(n) comparisons), the overall time complexity is O(n).
In summary, both operations individually have O(n) time complexity, and their combination results in an overall time complexity of O(n). Therefore, the statement is True.
