What should be size of map if different objects(say 3) have same hash code, and as a result, present in same bucket?

Question

What should be size of map if different objects(say 3) have same hash code, and as a result, present in same bucket?

Answer 1

The resulting size of the hash table depends on what collision resolution scheme we are using.

In the simplest case, we are using something like separate chaining (with linked lists).

In this case, we will have an array of N buckets and each bucket contains a reference to a linked list.

If we proceed to insert 3 items into the hash table, all of which share the same hash code, then the single target linked list would grow to length 3.

Thus, at a high level, we need at least N "units" of space to store bucket references plus 3 "units" of space to store the elements of the (occupied) linked list.

The exact size of these "units", depends on implementation details, such as word size (32-bit vs. 64-bit) and the exact definition of the linked list (singly- vs. doubly-linked list).

Assuming that we use singly-linked lists (for each bucket) on a 32-bit machine, the total size would be (approximately) 32 * N + (32 + x) * 3, where x refers to the size of the data type we are storing (e.g. ints, doubles, string, etc.)

If you would like to learn more, I would suggest googling "hash table collision" for more info.