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How can hashset.contains be O(1) with this implementation?

By : Siemer Family
Date : September 14 2020, 05:00 PM
wish helps you The classic implementation of a hash table works by assigning elements to one of a number of buckets, based on the hash of the element. If the hashing was perfect, i.e. no two elements had the same hash, then we'd be living in a perfectly perfect world where we wouldn't need to care about anything - any lookup would be O(1) always, because we'd only need to compute the hash, get the bucket and say if something is inside.
We're not living in a perfectly perfect world. First off, consider string hashing. In .NET, there are (2^16)^n possible strings of length n; GetHashCode returns a long, and there are 2^64 possible values of long. That's exactly enough to hash every string of length 4 to a unique long, but if we want strings longer than that, there must exist two different values that give the same hash - this is called a collision. Also, we don't want to maintain 2^64 buckets at all times anyway. The usual way of dealing with that is to take the hashcode and compute its value modulo the number of buckets to determine the bucket's number1. So, the takeaway is - we need to allow for collisions.
code :

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Null Object in HashSet implementation

By : Deval
Date : March 29 2020, 07:55 AM
this one helps. Because the HashSet contract specifies that remove() return true if the specified object existed and was removed. To do this, it uses the wrapped HashMap#remove() which returns the removed value.
If you were to store null instead of an object, then the call to HashMap#remove() would return null, which would be indistinguishable from the result of attempting to remove a non-existent object, and the contract of HashSet.remove() could not be fulfilled.

What would make this HashSet implementation fail?

By : Seletrix
Date : March 29 2020, 07:55 AM
hope this fix your issue In order to make my life easier dealing to strings, I want to use a hashset initialized with StringComparer.OrdinalIgnoreCase. ,
In what conditions isn't it safe?

HashSet internally uses hashmap for its implementation, then why is hashmap faster than hashset?

By : Sanjay Zala
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Because HashSet uses a HashMap. It must incur the cost of using a HashMap, plus the overhead of the HashSet itself.

Hashset implementation output

By : brooke
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further The add(..) function returns false if a string with the same contents was present in the HashSet (and the duplicate is obviously not added). Hence, the if (!s.add(a)) will be true, if duplicate strings are present in the args; s.size() <= args.length.

Using own implementation of HashSet.add()

By : Moataz ZS
Date : March 29 2020, 07:55 AM
it fixes the issue You should implement the equals and hashCode method in your User class.
As an example:
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