Question:
In search by hashing, we have size of the hash table and the size of list, then:
In search by hashing, we have size of the hash table and the size of list, then:
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In practice, the size of the hash table is often chosen to be larger than the size of the list to optimize search and reduce the occurrence of collisions.
Updated On: Apr 24, 2025
- Size of hash table is always smaller than size of list
- Size of hash table always equal to size of list
- Size of list is always greater than size of hashtable
- Size of hash table can be greater than the size of the list
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The Correct Option is D
Solution and Explanation
In hashing, the size of the hash table is not necessarily restricted to be smaller than the size of the list. In fact, it is common to have the hash table's size larger than the size of the list to minimize collisions and to maintain efficient performance. By using a larger hash table, we reduce the likelihood of multiple items being hashed to the same index. Therefore, the correct answer is "Size of hash table can be greater than the size of the list."
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Top Questions on Hashing
- All the elements that hash to the same slot are placed into the same linked list in:
- In ................. the search time is independent of the number of elements $n$.
- 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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