What is the next line? >>> tupleB = (5, 7, 5, 10, 2, 5) >>> tupleB. index(5,3) o5 o0 o1 o 3​

Instead of using a linked list to resolve collisions, as in separate chaining, use a binary search tree. that is, create a hash table that is an array of trees. to display a small tree-based hash table, you could use an inorder traversal of each tree. the advantage of a tree over a linked list is that it can be searched in o(logn) instead of o(n) time. this time savings can be a significant advantage if very high load factors are encountered. checking 15 items takes a maximum of 15 comparisons in a list but only 4 in a tree. duplicates can present problems in both trees and hash tables, so add some code that prevents a duplicate key from being inserted in the hash table. (beware: the find()method in tree assumes a non-empty tree.) to shorten the listing for this program, you can forget about deletion, which for trees requires a lot of