Answer:
i. A min heap of height 3 will have a root node with two children, each of which has two children of its own, resulting in a total of 7 nodes at the bottom level. Therefore:
The largest number of nodes that T can have is 1 + 2 + 4 + 7 = 14.
The smallest number of nodes that T can have is 1 + 2 + 4 = 7.
ii. The worst case complexity of deleting any arbitrary node element from a heap is O(log n), where n is the number of nodes in the heap. This is because deleting a node from a heap requires maintaining the heap property, which involves swapping the deleted node with its child nodes in order to ensure that the heap remains complete and that the heap property is satisfied. This process requires traversing the height of the tree, which has a worst-case complexity of O(log n).
Step-by-step explanation: