All the elements that hash to the same slot are placed into the same linked list in:
Show Hint
- Universal hashing
- Linear Probing
- Quadratic probing
- Chaining
The Correct Option is D
Solution and Explanation
Step 1: Understanding Hashing Methods.
- **Universal hashing**: This method involves choosing a hash function at random to avoid patterns. It does not use linked lists to resolve collisions.
- **Linear Probing**: This technique resolves collisions by finding the next available slot in a linear sequence, not by using linked lists.
- **Quadratic probing**: This is a collision resolution method that checks a quadratic sequence of positions to find an available slot, also without using linked lists.
- **Chaining**: Chaining handles collisions by storing all elements that hash to the same slot in a linked list. This is the correct method where linked lists are used for collision resolution.
Step 2: Conclusion.
Thus, the correct answer is **(4) Chaining**.
Top Questions on Hashing
- In ................. the search time is independent of the number of elements $n$.
- In search by hashing, we have size of the hash table and the size of list, then:
- Which of the following is(are) true for hashing
(A) The time required to compute the index value is independent of the number of items in the list
(B) The time required to compute the index value depend on the number of items in the list
(C) The cost of computing a hash function does not affect the performance of hashing based searching
(D) The time taken by different hash functions may be different
(E) There is no problem of collision in hashing
Choose the correct answer from the options given below : - Suppose we are given \(n\) keys, \(m\) hash table slots, and two simple uniform hash functions \(h_1\) and \(h_2\). Further suppose our hashing scheme uses \(h_1\) for the odd keys and \(h_2\) for the even keys. What is the expected number of keys in a slot?
Consider a dynamic hashing approach for 4-bit integer keys:
- There is a main hash table of size 4.
- The 2 least significant bits of a key are used to index into the main hash table.
- Initially, the main hash table entries are empty.
- To resolve collisions, the set of keys corresponding to a main hash table entry is organized as a binary tree that grows on demand.
- First, the 3rd least significant bit is used to divide the keys into left and right subtrees.
- Further collisions are resolved by subdividing based on the 4th least significant bit.
- A split is done only if needed, i.e., only when there is a collision.
Consider the following state of the hash table. Which of the following sequences of key insertions can cause the above state of the hash table (assume the keys are in decimal notation)?
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