Answer and Explanation:
1. In the case of the unordered list, the search algorithm is applied as the binary search would not be worked here also the time complexity for searching for n number of elements is O(n)
So the average number of nodes for the unordered list is n ÷ 2
2. For the ordered list, the same is to be considered i.e. n ÷ 2
Also it deals with the recepts of items and every item contains a relative position that represents an attribute of an item
3. For the unordered array, the same is to be considered i.e. n ÷ 2 but in this no sequence is followed
4. For the ordered array, the average number of nodes is [tex](log_2n) - 1[/tex] and in this, the binary search is to be considered
