Question

Write an algorithm to delete an element from a hash table that uses linear probing as its clash resolution strategy, Analyze your algorithm, and show the results using order notation.

Short answer

The algorithm starts by computing the hash of the key to be deleted. If the hashed index is empty, the key isn't in the table. If the key at the hashed index is the key to be deleted, it gets marked as a dummy node. If not, it steps linearly through the array until the key is found or an empty cell is reached. The algorithm then repeats this process until it has scanned the entire table once. The time complexity of this algorithm can be as quick as \(O(1)\) in the best case (no collisions), but can be as slow as \(O(n)\) in the worst case where n is the number of slots in the hash table.

Step-by-step solution

Understanding Hash Tables and Linear Probing

A hash table is a data structure that implements an associative array abstract data type, a structure that can map keys to values. A hash table uses a hash function to compute an index into an array of buckets or slots, from which the desired value can be found. When a collision occurs (two keys hash to the same index), Linear Probing handles it by probing, or searching through the table until it finds an empty spot where it shops the value.

Writing the Algorithm

Here is a primitive algorithm for deletion in a hash table using linear probing: 1. Compute the hash code of the key to be deleted, call this \(index\).2. If the hash table at index is empty, then the key we want to delete doesn't exist in the table. So, stop.3. If the key at index equals the key to be deleted, remove the key-value and mark it as deleted or a 'dummy' node. 4. If the key at index doesn't match the key to be deleted, keep moving in the array until either the key is found or an empty cell is found.5. Repeat the process until traversing the entire hash table once.

Analyzing the Algorithm

In the best-case scenario, the key to be deleted will be found at the initial index i.e., immediate success which gives the time complexity of \(O(1)\). This is the case where there are no collisions. However, in the worst case, there could be many collisions leading to a longer search time for an element in the hash table. Hence, in the worst case, it will go through all the cells of the hash table once which gives the time complexity of \(O(n)\), where n refers to the total number of slots in the hash table.