Find duplicate objects based on property in Java

Question

This is my object structure.

List<Customer> customerSelection = new ArrayList<Customer>();
    Customer c1 = new Customer();
    Customer c2 = new Customer();

    c1.setName("Syed");
    c2.setName("Syed");
    Map<String,String> locationList1 = new HashMap<String,String>();
    Map<String,String> locationList2 = new HashMap<String,String>();
    locationList1.put("DXB", "Dubai");
    locationList1.put("AUH", "Abudhabi");

    locationList2.put("DXB", "Dubai");
    locationList2.put("BAH", "Bahrain");

    c1.setLocationList(locationList1);
    c2.setLocationList(locationList2);

    customerSelection.add(c1);
    customerSelection.add(c2);

Here I want to validate if a customer is having preferred duplicate location, I should throw an error message. Any ideas on optimized solution?

Here Syed is having dubai as as location in location list which is invalid.

Why do you need a Map instead of a List for the locations? — Ayrton
– Ayrton, Commented Feb 14, 2019 at 12:37
@ayrton, this is an existing object structure, so I didn't bother about it which is having location id and location name. — Syed
– Syed, Commented Feb 14, 2019 at 12:38
I am not fully understand what you are asking. You don't want to duplicate key and value or you don't want to duplicate keys, or you don't want to duplicate values? — Said Alır
– Said Alır, Commented Feb 14, 2019 at 12:39
@SaidAlır, the customerSelection list will have duplicate objects based on customer name. But customer should not give duplicate preferred locations. — Syed
– Syed, Commented Feb 14, 2019 at 12:40
So, requirement is, they can have different names and same location, or same name and different locations, but never same name and same location? — weston
– weston, Commented Feb 14, 2019 at 12:52

weston · Accepted Answer · 2019-02-15 02:59:12Z

1

You can check the locations do not overlap at time of adding:

safeAdd(customerSelection, c2);

Using Collections.disjoint which returns true iff the two specified collections have no elements in common.

private static void safeAdd(List<Customer> customerSelection, Customer newCustomer) {
    if (customerSelection
            .stream()
            // look at all the users with matching name
            .filter(customer -> customer.name.equals(newCustomer.name))
            // ensure all of them have no location overlap with the new customer
            .allMatch(customer -> Collections.disjoint(customer.locationList.keySet(), newCustomer.locationList.keySet()))) {
        customerSelection.add(newCustomer);
    } else {
        throw new RuntimeException("Cannot add customer, customer with this name exists with this location");
    }
}

Or Java 7:

private static void safeAdd(List<Customer> customerSelection, Customer newCustomer) {
    for (Customer customer : customerSelection) {
        if (customer.name.equals(newCustomer.name) &&
                !Collections.disjoint(customer.locationList.keySet(), newCustomer.locationList.keySet())) {
            throw new RuntimeException("Cannot add customer, customer with this name exists with this location");
        }
    }
    customerSelection.add(newCustomer);
}

edited Feb 15, 2019 at 2:59

answered Feb 14, 2019 at 13:05

weston

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8 Comments

Syed Over a year ago

::( I'm having JDK as 1.7, so I can't use 1.8.., If possible please let me know in 1.7

weston Over a year ago

I'm sure you can work it out. Just use a for loop. Collections.disjoint is doing the hard work.

Syed Over a year ago

Thanks Weston. But how to call this method? while iteration of customerSelection? as I see you are passing safeAdd(customerSelection, c2); as hardcoded, because at the time of runtime, there can be multiple customers,right?

weston Over a year ago

We'll you call safeadd rather than add. I don't really understand what you don't understand.

Syed Over a year ago

My point is why are we passing c2 in this method as hardcoded.. safeAdd(customerSelection, c2); we just have to iterate the list and validate it right? Please correct me if Im wrong

|

Tejas Gundecha · Accepted Answer · 2019-02-14 13:53:26Z

0

You can iterate through list and check whether HashMap has same key present in other Customer's locationList.

for(int i=0; i<customerSelection.size()-1;i++){
   for(int j=0;j<customerSelection.size()-1;j++){
     if(j!=i){
   HashMap map   =    customerSelection.get(i).getLocationList();
   for(Map.Entry<> entry: map.entrySet()){
      if(customerSelection.get(j).getLocationList().containsKey(entry.getValue())) {
        THROW ERROR               
             }
      }

      }
   }

}

answered Feb 14, 2019 at 13:53

Tejas Gundecha

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2 Comments

Syed Over a year ago

for(Map.Entry<> entry: map.entrySet()){ .. This has compilation problem

Tejas Gundecha Over a year ago

Use Map.Entry<String,String>

Hakob Hakobyan · Accepted Answer · 2019-02-14 14:00:03Z

0

please have a look here too

    public static void main(String[] args) {
        List<Customer> customerSelection = new NonDuplicateLocationsList();
        Customer c1 = new Customer();
        Customer c2 = new Customer();

        c1.setName("Syed");
        c2.setName("Syed");
        Map<String,String> locationList1 = new HashMap<String,String>();
        Map<String,String> locationList2 = new HashMap<String,String>();
        locationList1.put("DXB", "Dubai");
        locationList1.put("AUH", "Abudhabi");

        locationList2.put("DXB", "Dubai");
        locationList2.put("BAH", "Bahrain");

        c1.setLocations(locationList1);
        c2.setLocations(locationList2);

        customerSelection.add(c1);
        customerSelection.add(c2);
    }
}
class Customer {
    Map<String, String> locations;
    String name;

    public Map<String, String> getLocations() {
        return locations;
    }

    public void setLocations(Map<String, String> locations) {
        this.locations = locations;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

class NonDuplicateLocationsList extends ArrayList<Customer> {


    @Override
    public boolean add(final Customer customer) {


        for (Customer customer1 : this) {
            if (Maps.difference(customer1.getLocations(),customer.getLocations()).entriesInCommon().size() > 0) {
                throw new RuntimeException("Has Location duplicates");
            }

        }
        return super.add(customer);
    }
}

edited Feb 14, 2019 at 14:00

answered Feb 14, 2019 at 13:18

Hakob Hakobyan

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4 Comments

Hakob Hakobyan Over a year ago

So I create a new class extended from Araylist, override add method and check all the stuff there, so that no one can add duplicate items there

Syed Over a year ago

The locationlist is a map object, but not list object. Please check it again.

Syed Over a year ago

But, please check it with a unit testing. Thanks Hakob

Hakob Hakobyan Over a year ago

@Syed I've updated the answer, tested for two cases, it works

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