I am making a custom implementation of a HashMap without using the HashMap data structure as part of an assignment, currently I have the choice of working with two 1D arrays or using a 2D array to store my keys and values. I want to be able to check if a key exists and return the corresponding value in O(1) time complexity (assignment requirement) but i am assuming it is without the use of containsKey().
Also, when inserting key and value pairs to my arrays, i am confused because it should not be O(1) logically, since there would occasionally be cases where there is collision and i have to recalculate the index, so why is the assignment requirement for insertion O(1)?
A lot of questions in there, let me give it a try.
I want to be able to check if a key exists and return the corresponding value in O(1) time complexity (assignment requirement) but i am assuming it is without the use of containsKey().
That actually doesn't make a difference. O(1) means the execution time is independent of the input, it does not mean a single operation is used. If your containsKey() and put() implementations are both O(1), then so is your solution that uses both of them exactly once.
Also, when inserting key and value pairs to my arrays, i am confused because it should not be O(1) logically, since there would occasionally be cases where there is collision and i have to recalculate the index, so why is the assignment requirement for insertion O(1)?
O(1) is the best case, which assumes that there are no hash collisions. The worst case is O(n) if every key generates the same hash code. So when a hash map's lookup or insertion performance is calculated as O(1), that assumes a perfect hashCode implementation.
Finally, when it comes to data structures, the usual approach is to use a single array, where the array items are link list nodes. The array offsets correspond to hashcode() % array size (there are much more advances formulas than this, but this is a good starting point). In case of a hash collision, you will have to navigate the linked list nodes until you find the correct entry.