Is there a data structure that only stores hash codes and not the actual objects?












7















My use-case is that I'm looking for a data structure in Java that will let me see if an object with the same hash code is inside (by calling contains()), but I will never need to iterate through the elements or retrieve the actual objects. A HashSet is close, but from my understanding, it still contains references to the actual objects, and that would be a waste of memory since I won't ever need the contents of the actual objects. The best option I can think of is a HashSet of type Integer storing only the hash codes, but I'm wondering if there is a built-in data structure that would accomplish the same thing (and only accept one type as opposed to HashSet of type Integer which will accept the hash code of any object).










share|improve this question




















  • 4





    Is your hash function perfect? Or can you have multiple objects with the same hash value?

    – arshajii
    4 hours ago






  • 6





    what about hashing collisions?

    – Nathan Hughes台湾不在中国
    4 hours ago






  • 7





    The HashSet will contain a reference to your object, not a copy, so don't worry about space. A HashSet<Integer> would probably use up more space because it has references to integers.

    – Sweeper
    4 hours ago













  • I agree with @Sweeper, unless you have a real need for super-duper optimization. Also, your second idea with storing hashcodes as integer wouln't be more efficient as it would store the hash+the hash of the hash.

    – Joel
    4 hours ago











  • @Sweeper The HashSet uses internally a HashMap. The memory space is the same.

    – Octavian R.
    4 hours ago


















7















My use-case is that I'm looking for a data structure in Java that will let me see if an object with the same hash code is inside (by calling contains()), but I will never need to iterate through the elements or retrieve the actual objects. A HashSet is close, but from my understanding, it still contains references to the actual objects, and that would be a waste of memory since I won't ever need the contents of the actual objects. The best option I can think of is a HashSet of type Integer storing only the hash codes, but I'm wondering if there is a built-in data structure that would accomplish the same thing (and only accept one type as opposed to HashSet of type Integer which will accept the hash code of any object).










share|improve this question




















  • 4





    Is your hash function perfect? Or can you have multiple objects with the same hash value?

    – arshajii
    4 hours ago






  • 6





    what about hashing collisions?

    – Nathan Hughes台湾不在中国
    4 hours ago






  • 7





    The HashSet will contain a reference to your object, not a copy, so don't worry about space. A HashSet<Integer> would probably use up more space because it has references to integers.

    – Sweeper
    4 hours ago













  • I agree with @Sweeper, unless you have a real need for super-duper optimization. Also, your second idea with storing hashcodes as integer wouln't be more efficient as it would store the hash+the hash of the hash.

    – Joel
    4 hours ago











  • @Sweeper The HashSet uses internally a HashMap. The memory space is the same.

    – Octavian R.
    4 hours ago
















7












7








7


1






My use-case is that I'm looking for a data structure in Java that will let me see if an object with the same hash code is inside (by calling contains()), but I will never need to iterate through the elements or retrieve the actual objects. A HashSet is close, but from my understanding, it still contains references to the actual objects, and that would be a waste of memory since I won't ever need the contents of the actual objects. The best option I can think of is a HashSet of type Integer storing only the hash codes, but I'm wondering if there is a built-in data structure that would accomplish the same thing (and only accept one type as opposed to HashSet of type Integer which will accept the hash code of any object).










share|improve this question
















My use-case is that I'm looking for a data structure in Java that will let me see if an object with the same hash code is inside (by calling contains()), but I will never need to iterate through the elements or retrieve the actual objects. A HashSet is close, but from my understanding, it still contains references to the actual objects, and that would be a waste of memory since I won't ever need the contents of the actual objects. The best option I can think of is a HashSet of type Integer storing only the hash codes, but I'm wondering if there is a built-in data structure that would accomplish the same thing (and only accept one type as opposed to HashSet of type Integer which will accept the hash code of any object).







java hashset






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago







B Yellow

















asked 4 hours ago









B YellowB Yellow

413




413








  • 4





    Is your hash function perfect? Or can you have multiple objects with the same hash value?

    – arshajii
    4 hours ago






  • 6





    what about hashing collisions?

    – Nathan Hughes台湾不在中国
    4 hours ago






  • 7





    The HashSet will contain a reference to your object, not a copy, so don't worry about space. A HashSet<Integer> would probably use up more space because it has references to integers.

    – Sweeper
    4 hours ago













  • I agree with @Sweeper, unless you have a real need for super-duper optimization. Also, your second idea with storing hashcodes as integer wouln't be more efficient as it would store the hash+the hash of the hash.

    – Joel
    4 hours ago











  • @Sweeper The HashSet uses internally a HashMap. The memory space is the same.

    – Octavian R.
    4 hours ago
















  • 4





    Is your hash function perfect? Or can you have multiple objects with the same hash value?

    – arshajii
    4 hours ago






  • 6





    what about hashing collisions?

    – Nathan Hughes台湾不在中国
    4 hours ago






  • 7





    The HashSet will contain a reference to your object, not a copy, so don't worry about space. A HashSet<Integer> would probably use up more space because it has references to integers.

    – Sweeper
    4 hours ago













  • I agree with @Sweeper, unless you have a real need for super-duper optimization. Also, your second idea with storing hashcodes as integer wouln't be more efficient as it would store the hash+the hash of the hash.

    – Joel
    4 hours ago











  • @Sweeper The HashSet uses internally a HashMap. The memory space is the same.

    – Octavian R.
    4 hours ago










4




4





Is your hash function perfect? Or can you have multiple objects with the same hash value?

– arshajii
4 hours ago





Is your hash function perfect? Or can you have multiple objects with the same hash value?

– arshajii
4 hours ago




6




6





what about hashing collisions?

– Nathan Hughes台湾不在中国
4 hours ago





what about hashing collisions?

– Nathan Hughes台湾不在中国
4 hours ago




7




7





The HashSet will contain a reference to your object, not a copy, so don't worry about space. A HashSet<Integer> would probably use up more space because it has references to integers.

– Sweeper
4 hours ago







The HashSet will contain a reference to your object, not a copy, so don't worry about space. A HashSet<Integer> would probably use up more space because it has references to integers.

– Sweeper
4 hours ago















I agree with @Sweeper, unless you have a real need for super-duper optimization. Also, your second idea with storing hashcodes as integer wouln't be more efficient as it would store the hash+the hash of the hash.

– Joel
4 hours ago





I agree with @Sweeper, unless you have a real need for super-duper optimization. Also, your second idea with storing hashcodes as integer wouln't be more efficient as it would store the hash+the hash of the hash.

– Joel
4 hours ago













@Sweeper The HashSet uses internally a HashMap. The memory space is the same.

– Octavian R.
4 hours ago







@Sweeper The HashSet uses internally a HashMap. The memory space is the same.

– Octavian R.
4 hours ago














4 Answers
4






active

oldest

votes


















8














A Bloom filter can tell whether an object might be a member, or is definitely not a member. You can control the likelihood of false positives. A single bit is stored per hash value.



The Guava library provides an implementation in Java.






share|improve this answer


























  • Nice. This seems like the solution for very low storage overhead. But you have to worry about the false negative case. If you can statistically eliminate that, this is great!

    – Steve
    4 hours ago








  • 1





    False positives, but you can control their probability. Another disadvantage is that you can't remove elements.

    – Andy Thomas
    4 hours ago











  • The question was for a data structure that checks using only the predefined hashCode(), which can potentially have 2^31 values (counting only positives). A Bloom filter that uses one hash function with 2 ^ 31 possible values would be extraordinarily large, seeing as it is basically just a BitSet. I don't see how that counts as "very low storage overhead".

    – Leo Aso
    3 hours ago





















1














If you want to track if a hash code is already present and to do it memory efficient a BitSet may suite your requirements.



Look at the following example:



  public static void main(String args) {
BitSet hashCodes = new BitSet();
hashCodes.set("1".hashCode());

System.out.println(hashCodes.get("1".hashCode())); // true
System.out.println(hashCodes.get("2".hashCode())); // false
}


The BitSet "implements a vector of bits that grows as needed.". It's a JDK "built-in data structure" which doesn't contain "references to the actual objects". It stores only if "the same hash code is inside".



EDIT:

As @Steve mentioned in his comment the implementation of the BitSet isn't the most memory efficient one. But there are more memory efficient implementations of a bit set - though not built-in.






share|improve this answer


























  • I don't know how a BitSet stores individual bits. Since obviously this usage will spread bits across a very large input domain, are you sure those are stored efficiently? Just asking. The naive assumption is that the structure would be an array of bytes, where the array was just expanded to include any bit position and all positions before that, which would be monstrously inefficient. But I don't know how it actually represents bits spread way apart.

    – Steve
    3 hours ago











  • It appears that your solution won't work. See github.com/brettwooldridge/SparseBitSet

    – Steve
    3 hours ago











  • @Steve You're right. Found additionally this post. But the idea of an bit set is basically not bad. It's rather the implementation of the JDK BitSet.

    – LuCio
    3 hours ago





















0














You could use a primitive collection implementation like IntSet to store values of hash codes. Obviously as others have mentioned this assumes collisions aren't a problem.






share|improve this answer































    -1














    There is no such built-in data structure, because such a data structure is rarely needed. It's easy to build one, though.



    public class HashCodeSet<T> {

    private final HashSet<Integer> hashCodes;

    public MyHashSet() {
    hashCodes = new HashSet<>();
    }

    public MyHashSet(int initialCapacity) {
    hashCodes = new HashSet<>(initialCapacity);
    }

    public HashCodeSet(HashCodeSet toCopy) {
    hashCodes = new HashSet<>(toCopy.hashCodes);
    }

    public void add(T element) {
    hashCodes.add(element.hashCode());
    }

    public boolean containsHashCodeOf(T element) {
    return hashCodes.contains(element.hashCode());
    }

    @Override
    public boolean equals(o: Object) {
    return o == this || o instanceof HashCodeSet &&
    ((HashCodeSet) o).hashCodes.equals(hashCodes);
    }

    @Override
    public int hashCode() {
    return hashCodes.hashCode(); // hash-ception
    }

    @Override
    public String toString() {
    return hashCodes.toString();
    }
    }





    share|improve this answer



















    • 1





      I think the OP's question wasn't about API, but about memory usage. This doesn't help with that since the result still acts like a HashSet.

      – Steve
      4 hours ago











    • I realize that I was unfair completely on this. I've removed my objections. Feel free to delete your comments on my comments. I still think there's something more to the problem, but I was off base. Sorry

      – Steve
      3 hours ago











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8














    A Bloom filter can tell whether an object might be a member, or is definitely not a member. You can control the likelihood of false positives. A single bit is stored per hash value.



    The Guava library provides an implementation in Java.






    share|improve this answer


























    • Nice. This seems like the solution for very low storage overhead. But you have to worry about the false negative case. If you can statistically eliminate that, this is great!

      – Steve
      4 hours ago








    • 1





      False positives, but you can control their probability. Another disadvantage is that you can't remove elements.

      – Andy Thomas
      4 hours ago











    • The question was for a data structure that checks using only the predefined hashCode(), which can potentially have 2^31 values (counting only positives). A Bloom filter that uses one hash function with 2 ^ 31 possible values would be extraordinarily large, seeing as it is basically just a BitSet. I don't see how that counts as "very low storage overhead".

      – Leo Aso
      3 hours ago


















    8














    A Bloom filter can tell whether an object might be a member, or is definitely not a member. You can control the likelihood of false positives. A single bit is stored per hash value.



    The Guava library provides an implementation in Java.






    share|improve this answer


























    • Nice. This seems like the solution for very low storage overhead. But you have to worry about the false negative case. If you can statistically eliminate that, this is great!

      – Steve
      4 hours ago








    • 1





      False positives, but you can control their probability. Another disadvantage is that you can't remove elements.

      – Andy Thomas
      4 hours ago











    • The question was for a data structure that checks using only the predefined hashCode(), which can potentially have 2^31 values (counting only positives). A Bloom filter that uses one hash function with 2 ^ 31 possible values would be extraordinarily large, seeing as it is basically just a BitSet. I don't see how that counts as "very low storage overhead".

      – Leo Aso
      3 hours ago
















    8












    8








    8







    A Bloom filter can tell whether an object might be a member, or is definitely not a member. You can control the likelihood of false positives. A single bit is stored per hash value.



    The Guava library provides an implementation in Java.






    share|improve this answer















    A Bloom filter can tell whether an object might be a member, or is definitely not a member. You can control the likelihood of false positives. A single bit is stored per hash value.



    The Guava library provides an implementation in Java.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 4 hours ago

























    answered 4 hours ago









    Andy ThomasAndy Thomas

    68k980133




    68k980133













    • Nice. This seems like the solution for very low storage overhead. But you have to worry about the false negative case. If you can statistically eliminate that, this is great!

      – Steve
      4 hours ago








    • 1





      False positives, but you can control their probability. Another disadvantage is that you can't remove elements.

      – Andy Thomas
      4 hours ago











    • The question was for a data structure that checks using only the predefined hashCode(), which can potentially have 2^31 values (counting only positives). A Bloom filter that uses one hash function with 2 ^ 31 possible values would be extraordinarily large, seeing as it is basically just a BitSet. I don't see how that counts as "very low storage overhead".

      – Leo Aso
      3 hours ago





















    • Nice. This seems like the solution for very low storage overhead. But you have to worry about the false negative case. If you can statistically eliminate that, this is great!

      – Steve
      4 hours ago








    • 1





      False positives, but you can control their probability. Another disadvantage is that you can't remove elements.

      – Andy Thomas
      4 hours ago











    • The question was for a data structure that checks using only the predefined hashCode(), which can potentially have 2^31 values (counting only positives). A Bloom filter that uses one hash function with 2 ^ 31 possible values would be extraordinarily large, seeing as it is basically just a BitSet. I don't see how that counts as "very low storage overhead".

      – Leo Aso
      3 hours ago



















    Nice. This seems like the solution for very low storage overhead. But you have to worry about the false negative case. If you can statistically eliminate that, this is great!

    – Steve
    4 hours ago







    Nice. This seems like the solution for very low storage overhead. But you have to worry about the false negative case. If you can statistically eliminate that, this is great!

    – Steve
    4 hours ago






    1




    1





    False positives, but you can control their probability. Another disadvantage is that you can't remove elements.

    – Andy Thomas
    4 hours ago





    False positives, but you can control their probability. Another disadvantage is that you can't remove elements.

    – Andy Thomas
    4 hours ago













    The question was for a data structure that checks using only the predefined hashCode(), which can potentially have 2^31 values (counting only positives). A Bloom filter that uses one hash function with 2 ^ 31 possible values would be extraordinarily large, seeing as it is basically just a BitSet. I don't see how that counts as "very low storage overhead".

    – Leo Aso
    3 hours ago







    The question was for a data structure that checks using only the predefined hashCode(), which can potentially have 2^31 values (counting only positives). A Bloom filter that uses one hash function with 2 ^ 31 possible values would be extraordinarily large, seeing as it is basically just a BitSet. I don't see how that counts as "very low storage overhead".

    – Leo Aso
    3 hours ago















    1














    If you want to track if a hash code is already present and to do it memory efficient a BitSet may suite your requirements.



    Look at the following example:



      public static void main(String args) {
    BitSet hashCodes = new BitSet();
    hashCodes.set("1".hashCode());

    System.out.println(hashCodes.get("1".hashCode())); // true
    System.out.println(hashCodes.get("2".hashCode())); // false
    }


    The BitSet "implements a vector of bits that grows as needed.". It's a JDK "built-in data structure" which doesn't contain "references to the actual objects". It stores only if "the same hash code is inside".



    EDIT:

    As @Steve mentioned in his comment the implementation of the BitSet isn't the most memory efficient one. But there are more memory efficient implementations of a bit set - though not built-in.






    share|improve this answer


























    • I don't know how a BitSet stores individual bits. Since obviously this usage will spread bits across a very large input domain, are you sure those are stored efficiently? Just asking. The naive assumption is that the structure would be an array of bytes, where the array was just expanded to include any bit position and all positions before that, which would be monstrously inefficient. But I don't know how it actually represents bits spread way apart.

      – Steve
      3 hours ago











    • It appears that your solution won't work. See github.com/brettwooldridge/SparseBitSet

      – Steve
      3 hours ago











    • @Steve You're right. Found additionally this post. But the idea of an bit set is basically not bad. It's rather the implementation of the JDK BitSet.

      – LuCio
      3 hours ago


















    1














    If you want to track if a hash code is already present and to do it memory efficient a BitSet may suite your requirements.



    Look at the following example:



      public static void main(String args) {
    BitSet hashCodes = new BitSet();
    hashCodes.set("1".hashCode());

    System.out.println(hashCodes.get("1".hashCode())); // true
    System.out.println(hashCodes.get("2".hashCode())); // false
    }


    The BitSet "implements a vector of bits that grows as needed.". It's a JDK "built-in data structure" which doesn't contain "references to the actual objects". It stores only if "the same hash code is inside".



    EDIT:

    As @Steve mentioned in his comment the implementation of the BitSet isn't the most memory efficient one. But there are more memory efficient implementations of a bit set - though not built-in.






    share|improve this answer


























    • I don't know how a BitSet stores individual bits. Since obviously this usage will spread bits across a very large input domain, are you sure those are stored efficiently? Just asking. The naive assumption is that the structure would be an array of bytes, where the array was just expanded to include any bit position and all positions before that, which would be monstrously inefficient. But I don't know how it actually represents bits spread way apart.

      – Steve
      3 hours ago











    • It appears that your solution won't work. See github.com/brettwooldridge/SparseBitSet

      – Steve
      3 hours ago











    • @Steve You're right. Found additionally this post. But the idea of an bit set is basically not bad. It's rather the implementation of the JDK BitSet.

      – LuCio
      3 hours ago
















    1












    1








    1







    If you want to track if a hash code is already present and to do it memory efficient a BitSet may suite your requirements.



    Look at the following example:



      public static void main(String args) {
    BitSet hashCodes = new BitSet();
    hashCodes.set("1".hashCode());

    System.out.println(hashCodes.get("1".hashCode())); // true
    System.out.println(hashCodes.get("2".hashCode())); // false
    }


    The BitSet "implements a vector of bits that grows as needed.". It's a JDK "built-in data structure" which doesn't contain "references to the actual objects". It stores only if "the same hash code is inside".



    EDIT:

    As @Steve mentioned in his comment the implementation of the BitSet isn't the most memory efficient one. But there are more memory efficient implementations of a bit set - though not built-in.






    share|improve this answer















    If you want to track if a hash code is already present and to do it memory efficient a BitSet may suite your requirements.



    Look at the following example:



      public static void main(String args) {
    BitSet hashCodes = new BitSet();
    hashCodes.set("1".hashCode());

    System.out.println(hashCodes.get("1".hashCode())); // true
    System.out.println(hashCodes.get("2".hashCode())); // false
    }


    The BitSet "implements a vector of bits that grows as needed.". It's a JDK "built-in data structure" which doesn't contain "references to the actual objects". It stores only if "the same hash code is inside".



    EDIT:

    As @Steve mentioned in his comment the implementation of the BitSet isn't the most memory efficient one. But there are more memory efficient implementations of a bit set - though not built-in.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 3 hours ago

























    answered 4 hours ago









    LuCioLuCio

    2,8821924




    2,8821924













    • I don't know how a BitSet stores individual bits. Since obviously this usage will spread bits across a very large input domain, are you sure those are stored efficiently? Just asking. The naive assumption is that the structure would be an array of bytes, where the array was just expanded to include any bit position and all positions before that, which would be monstrously inefficient. But I don't know how it actually represents bits spread way apart.

      – Steve
      3 hours ago











    • It appears that your solution won't work. See github.com/brettwooldridge/SparseBitSet

      – Steve
      3 hours ago











    • @Steve You're right. Found additionally this post. But the idea of an bit set is basically not bad. It's rather the implementation of the JDK BitSet.

      – LuCio
      3 hours ago





















    • I don't know how a BitSet stores individual bits. Since obviously this usage will spread bits across a very large input domain, are you sure those are stored efficiently? Just asking. The naive assumption is that the structure would be an array of bytes, where the array was just expanded to include any bit position and all positions before that, which would be monstrously inefficient. But I don't know how it actually represents bits spread way apart.

      – Steve
      3 hours ago











    • It appears that your solution won't work. See github.com/brettwooldridge/SparseBitSet

      – Steve
      3 hours ago











    • @Steve You're right. Found additionally this post. But the idea of an bit set is basically not bad. It's rather the implementation of the JDK BitSet.

      – LuCio
      3 hours ago



















    I don't know how a BitSet stores individual bits. Since obviously this usage will spread bits across a very large input domain, are you sure those are stored efficiently? Just asking. The naive assumption is that the structure would be an array of bytes, where the array was just expanded to include any bit position and all positions before that, which would be monstrously inefficient. But I don't know how it actually represents bits spread way apart.

    – Steve
    3 hours ago





    I don't know how a BitSet stores individual bits. Since obviously this usage will spread bits across a very large input domain, are you sure those are stored efficiently? Just asking. The naive assumption is that the structure would be an array of bytes, where the array was just expanded to include any bit position and all positions before that, which would be monstrously inefficient. But I don't know how it actually represents bits spread way apart.

    – Steve
    3 hours ago













    It appears that your solution won't work. See github.com/brettwooldridge/SparseBitSet

    – Steve
    3 hours ago





    It appears that your solution won't work. See github.com/brettwooldridge/SparseBitSet

    – Steve
    3 hours ago













    @Steve You're right. Found additionally this post. But the idea of an bit set is basically not bad. It's rather the implementation of the JDK BitSet.

    – LuCio
    3 hours ago







    @Steve You're right. Found additionally this post. But the idea of an bit set is basically not bad. It's rather the implementation of the JDK BitSet.

    – LuCio
    3 hours ago













    0














    You could use a primitive collection implementation like IntSet to store values of hash codes. Obviously as others have mentioned this assumes collisions aren't a problem.






    share|improve this answer




























      0














      You could use a primitive collection implementation like IntSet to store values of hash codes. Obviously as others have mentioned this assumes collisions aren't a problem.






      share|improve this answer


























        0












        0








        0







        You could use a primitive collection implementation like IntSet to store values of hash codes. Obviously as others have mentioned this assumes collisions aren't a problem.






        share|improve this answer













        You could use a primitive collection implementation like IntSet to store values of hash codes. Obviously as others have mentioned this assumes collisions aren't a problem.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 3 hours ago









        MarkMark

        24.2k44783




        24.2k44783























            -1














            There is no such built-in data structure, because such a data structure is rarely needed. It's easy to build one, though.



            public class HashCodeSet<T> {

            private final HashSet<Integer> hashCodes;

            public MyHashSet() {
            hashCodes = new HashSet<>();
            }

            public MyHashSet(int initialCapacity) {
            hashCodes = new HashSet<>(initialCapacity);
            }

            public HashCodeSet(HashCodeSet toCopy) {
            hashCodes = new HashSet<>(toCopy.hashCodes);
            }

            public void add(T element) {
            hashCodes.add(element.hashCode());
            }

            public boolean containsHashCodeOf(T element) {
            return hashCodes.contains(element.hashCode());
            }

            @Override
            public boolean equals(o: Object) {
            return o == this || o instanceof HashCodeSet &&
            ((HashCodeSet) o).hashCodes.equals(hashCodes);
            }

            @Override
            public int hashCode() {
            return hashCodes.hashCode(); // hash-ception
            }

            @Override
            public String toString() {
            return hashCodes.toString();
            }
            }





            share|improve this answer



















            • 1





              I think the OP's question wasn't about API, but about memory usage. This doesn't help with that since the result still acts like a HashSet.

              – Steve
              4 hours ago











            • I realize that I was unfair completely on this. I've removed my objections. Feel free to delete your comments on my comments. I still think there's something more to the problem, but I was off base. Sorry

              – Steve
              3 hours ago
















            -1














            There is no such built-in data structure, because such a data structure is rarely needed. It's easy to build one, though.



            public class HashCodeSet<T> {

            private final HashSet<Integer> hashCodes;

            public MyHashSet() {
            hashCodes = new HashSet<>();
            }

            public MyHashSet(int initialCapacity) {
            hashCodes = new HashSet<>(initialCapacity);
            }

            public HashCodeSet(HashCodeSet toCopy) {
            hashCodes = new HashSet<>(toCopy.hashCodes);
            }

            public void add(T element) {
            hashCodes.add(element.hashCode());
            }

            public boolean containsHashCodeOf(T element) {
            return hashCodes.contains(element.hashCode());
            }

            @Override
            public boolean equals(o: Object) {
            return o == this || o instanceof HashCodeSet &&
            ((HashCodeSet) o).hashCodes.equals(hashCodes);
            }

            @Override
            public int hashCode() {
            return hashCodes.hashCode(); // hash-ception
            }

            @Override
            public String toString() {
            return hashCodes.toString();
            }
            }





            share|improve this answer



















            • 1





              I think the OP's question wasn't about API, but about memory usage. This doesn't help with that since the result still acts like a HashSet.

              – Steve
              4 hours ago











            • I realize that I was unfair completely on this. I've removed my objections. Feel free to delete your comments on my comments. I still think there's something more to the problem, but I was off base. Sorry

              – Steve
              3 hours ago














            -1












            -1








            -1







            There is no such built-in data structure, because such a data structure is rarely needed. It's easy to build one, though.



            public class HashCodeSet<T> {

            private final HashSet<Integer> hashCodes;

            public MyHashSet() {
            hashCodes = new HashSet<>();
            }

            public MyHashSet(int initialCapacity) {
            hashCodes = new HashSet<>(initialCapacity);
            }

            public HashCodeSet(HashCodeSet toCopy) {
            hashCodes = new HashSet<>(toCopy.hashCodes);
            }

            public void add(T element) {
            hashCodes.add(element.hashCode());
            }

            public boolean containsHashCodeOf(T element) {
            return hashCodes.contains(element.hashCode());
            }

            @Override
            public boolean equals(o: Object) {
            return o == this || o instanceof HashCodeSet &&
            ((HashCodeSet) o).hashCodes.equals(hashCodes);
            }

            @Override
            public int hashCode() {
            return hashCodes.hashCode(); // hash-ception
            }

            @Override
            public String toString() {
            return hashCodes.toString();
            }
            }





            share|improve this answer













            There is no such built-in data structure, because such a data structure is rarely needed. It's easy to build one, though.



            public class HashCodeSet<T> {

            private final HashSet<Integer> hashCodes;

            public MyHashSet() {
            hashCodes = new HashSet<>();
            }

            public MyHashSet(int initialCapacity) {
            hashCodes = new HashSet<>(initialCapacity);
            }

            public HashCodeSet(HashCodeSet toCopy) {
            hashCodes = new HashSet<>(toCopy.hashCodes);
            }

            public void add(T element) {
            hashCodes.add(element.hashCode());
            }

            public boolean containsHashCodeOf(T element) {
            return hashCodes.contains(element.hashCode());
            }

            @Override
            public boolean equals(o: Object) {
            return o == this || o instanceof HashCodeSet &&
            ((HashCodeSet) o).hashCodes.equals(hashCodes);
            }

            @Override
            public int hashCode() {
            return hashCodes.hashCode(); // hash-ception
            }

            @Override
            public String toString() {
            return hashCodes.toString();
            }
            }






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 4 hours ago









            Leo AsoLeo Aso

            5,27211029




            5,27211029








            • 1





              I think the OP's question wasn't about API, but about memory usage. This doesn't help with that since the result still acts like a HashSet.

              – Steve
              4 hours ago











            • I realize that I was unfair completely on this. I've removed my objections. Feel free to delete your comments on my comments. I still think there's something more to the problem, but I was off base. Sorry

              – Steve
              3 hours ago














            • 1





              I think the OP's question wasn't about API, but about memory usage. This doesn't help with that since the result still acts like a HashSet.

              – Steve
              4 hours ago











            • I realize that I was unfair completely on this. I've removed my objections. Feel free to delete your comments on my comments. I still think there's something more to the problem, but I was off base. Sorry

              – Steve
              3 hours ago








            1




            1





            I think the OP's question wasn't about API, but about memory usage. This doesn't help with that since the result still acts like a HashSet.

            – Steve
            4 hours ago





            I think the OP's question wasn't about API, but about memory usage. This doesn't help with that since the result still acts like a HashSet.

            – Steve
            4 hours ago













            I realize that I was unfair completely on this. I've removed my objections. Feel free to delete your comments on my comments. I still think there's something more to the problem, but I was off base. Sorry

            – Steve
            3 hours ago





            I realize that I was unfair completely on this. I've removed my objections. Feel free to delete your comments on my comments. I still think there's something more to the problem, but I was off base. Sorry

            – Steve
            3 hours ago


















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