Instead of using an array to store pointers to linked lists, you would like to explore using an AVL tree that you have pre-populated with a node for each address in the hash table. Therefore, the number of nodes n in the AVL tree corresponds to the size of the hash table. How would using an AVL tree to store the pointers to the linked lists affect the running time of insertions, deletions, and finds in the hash table? Your answer must include a justification that describes the running time in terms of the hash table size. You must use big-Oh notation. For clarification, here is an example of a hash table (with table size = 6) using Separate Chaining. "Prima" "Butera" %3D "Smith" %3D 1 2 "Presley" / 3 "Sinatra" "Martin" "Davis" 5 Here is an example of the same items stored in the new AVL Tree – Separate Chaining approach you must explore. 3 1 5 "Sinatra" "Martin" "Davis" "Presley" / "Prima" "Butera" "Smith" /

How many linked lists are there in a chaining hash table of size T? What is/are the elementary data structure(s) used in Separate Chaining hash table?

Implement a commonly used hash table in a program that handles collision using linear probing. Using (K mod 13) as the hash function, store the following elements in the table: {1, 5, 21, 26, 39, 14, 15, 16, 17, 18, 19, 20, 111, 145, 146}.

Can m-linked-lists be used to create a hash table? Why do we need a hash function, exactly? Provide an illustration.

Suppose you have a hash table of size N = 64, and you are using pseudo-random probing. The keys in your hash are 4-digit integers (0000 through 9999) and your hash function is h(k) = (the sum of the digits in k). Assume that the first 4 slots of your pseudo-random probing array contain: 5, 10, 60, 30 What are the first 4 values in the probe sequence (starting with the home position) for a record with key k=1948?

Suppose you have a hash table with seven entries (indexed 0 through 6). This table uses open addressing with the hash function that maps each letter to its alphabet code (a = A = 0, etc.) modulo 7. Rehashing is accomplished using linear-probing with a jump of 1. Describe the state of the table after each of the letters D, a, d, H, a, and h are added to the table.

What is the worst-case performance of a lookup operation in a hashmap and why? Group of answer choices A- O(1), hashmap always has a constant time lookup, and that is why we like using this associative data structure.   B- O(lg(n)) hashmap has a log(n) lookup because we are able to perform a binary search on the keys because our hashmap always maintains a sorted order of entries added.   C- O(n) because we can have a bad hash function that puts all of our items in the same bucket, thus we would have to iterate through all n items.

What would happen to insertions, deletions, and searches in the hash table if the linked list pointers were preserved in an AVL tree instead of the traditional tree format? In your answer, you are required to provide an explanation of the running time based on the size of the hash table. Is the Big-Ohnotation really required?

What is the time complexity for appending an item to a hash table with Open Addressing in worst case? Explain.

Does a hashing of size m always contain the same number of linked lists? Hash functions make no sense to me at all. Provide an illustration to back up your claims.