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Trying to create a RetainAll method

I am trying to create a RetainAll method and all after scouring the forums I have found nothing that helps in my specific case. The issue I am having is that when running my program instead of retaining all the elements form a specified list and deleting all others it instead keeps the last element form the initial list.
public default boolean retainAll(Collection<?> c) {
boolean modified=false;
int index =0;
for(Object e : this) {
if(c.contains(e)==true) {
}
else if(c.contains(e)==false) {
index=this.indexOf(e);
this.remove(index);
modified = true;
}
}
return modified;
}
public default boolean remove(Object e) {
if (indexOf(e) >= 0) {
remove(indexOf(e));
return true;
}
else
return false;
}
I am just not understanding how to remove the last element.

You're modifying the collection as you iterate over it. That is generally a recipe for unhappiness.
For the standard collections, the safe way is to use an explicit iterator, and use the iterator's 'remove' method to remove the most recently returned entry.
If this is your own collection implementation you should make sure that model works for you too.

Comparing raw object with a Stack/Queue - Overriding equals() method

I've created a class which can be inherited to create both a Stack and a Queue using LinkedLists, I've passed all the JUnit tests except the equals one, I still have no idea why it doesn't work.
#Override public boolean equals(Object o) {
if( o == null) return false;
if(o == this) return true;
if(!(o instanceof PushPop)) return false;
PushPop test1= this;
PushPop test = (PushPop)o;
while(!test.isEmpty() && !test1.isEmpty()){
if(test1.pop() != test.pop()) return false;
}
return true;
}
The test sends out an assertion error whenever it's comparing the values, specifically whenever a stack/queue has an extra value than the second one.

OK, I have located your problem:
In your test, you do the following [taken from comment]
stack.push(i);
Assert.assertFalse(stack.equals(stack2));
stack2.push(i);
Assert.assertTrue(stack.equals(stack2));
This seems reasonable. However, equals clears both stacks so when you push i onto stack2, it is no longer equal to the now-empty stack.
Hence your error.
The solution: Don't Modify Your Object In An equals Method.
I'd suggest cloning or comparing whatever your underlying data structure is (e.g., if you're using Nodes, having Node implement equals).

I have two arrayList<myObject>, I'm not able to use equals() to be able to compare them correctly. Why?

I have two arrayLists<myObject>, where myObject is an object of a custom class I've created. I want to be able to compare those arrayLists using the equals() method.
After reading and looking for answers, I've read that certain objects like int[] are only considered equal by the equals() method when they are referencing the same thing.
To fix that, I tried to override the equals method in my custom object. My objects have 3 atributes (all basic types), so my equals method now returns true if all the 3 atributes are equal to those of the object compared, and false otherwise. However, comparing the arraylists still doesn't work. What am I doing wrong?
Excuse me for explaining the code instead of posting it, I do it because the variables and names aren't in English.
EDIT: Ok, here's the code. Compra is my custom class; cantidad,concepto and id are its atributes.
#Override
public boolean equals(Object obj) {
boolean result = true;
if (obj == null) {
result = false;
}else{
Compra comprobada = (Compra) obj;
if(!(this.id == comprobada.getId())){
result = false;
}
if(!(this.cantidad == comprobada.getCantidad())){
result = false;
} if(!this.concepto.equals(comprobada.getConcepto())){
result = false;
}
}
return result;
}

Based on this one :
How can I check if two ArrayList differ, I don't care what's changed
If you have implemented your custom object equals correct (you actually override it and have your one) and the size of the arrayList is the same and each of the pair of the objects is equal then it will return equal. In other words what you are trying to do is totally correct but your arrayLists are not actually having exactly the equal objects in exact order.
Make sure that your equal is called when you check for collection equality by doing a System.out.println() to investigate what is going on.
If you don't mind please send the equals of your object.
I run your code in an isolated example and works fine (outtputs true) - I improved the equals method so it doesn't do so many if checks as if only one of them is not equal it should return false.
class stackoverflow {
public static void main(String args[]){
ArrayList<Compra> array1 = new ArrayList<>();
ArrayList<Compra> array2 = new ArrayList<>();
array1.add(new Compra(1,2,"test"));
array2.add(new Compra(1,2,"test"));
System.out.println(array1.equals(array2));
}
}
class Compra {
int id;
int cantidad;
String concepto;
public Compra(int id, int cantidad, String concepto){
this.id = id;
this.cantidad = cantidad;
this.concepto = concepto;
}
public boolean equals(Object obj) {
if (obj == null) {
return false;
}else{
Compra comprobada = (Compra) obj;
if(!(this.id == comprobada.getId())){
return false;
}
if(!(this.cantidad == comprobada.getCantidad())){
return false;
}
if(!this.concepto.equals(comprobada.getConcepto())){
return false;
}
}
return true;
}
public int getId() {
return id;
}
public int getCantidad() {
return cantidad;
}
public String getConcepto() {
return concepto;
}
}
Some things to check:
Are you sure you don't change the order of the things in ArrayList??:
Do you print to make sure that these equals checks happen and return true or false as expected?
Are you sure that concepto Strings are exactly the same, with the same case and don't contain extra spaces etc?

As you haven't posted code i suggest you to check into Comparable class and method compareTo and how to use it for custom classes.

DFS tree traversal function modification

Please find below my implementation for DFS.
protected void DFS(String search) {
for(Tree<T> child : leafs) {
if(child.value.equals(search))
return;
else
child.DFS(search);
System.out.println(child.value);
}
}
The objective is to stop traversal on finding the node whose value is in the variable search. However, the above function goes on traversing the tree even beyond the declared search node. Could someone help me modify the above function?
Thank you.
Edit 1
protected boolean DFS(String anaphorKey) {
boolean found = false;
for(Tree<T> child : leafs) {
if(child.head.equals(anaphorKey))
return true;
found = child.DFS(anaphorKey);
if(found == true)
break;
System.out.println(child.head);
//System.out.println("anaphorKey: "+anaphorKey);
}
return found;
}
Tried implementing the given answer suggestion (#SJuan76). The implementation above isn't working as desired. Could you point me to the place where code is not as per the logic suggested?

rookie, might I suggest an implementation using the classic for-loop (as opposed to the enhanced for-loop being used now) which allows integration of your stop-condition a bit better, something like:
protected boolean DFS(String key) {
boolean found = false;
for(int i = 0; i < leafs.size() && !found; i++) {
Tree<T> child = leafs.get(i);
if(child.head.equals(key))
found = true;
else
found = child.DFS(key);
}
return found;
}
So as soon as your found condition is hit, the 'found' becomes true and your loop stops.
What you may have forgotten is the "found = child.DFS(key)" portion of the recursion, where you need to remember the result of your recursive calls so ALL your for-loops on up the chain all break as soon as you return.
Hope that helps.

Option A (Nice): the function returns a value, when the node is found it returns a different value that if the node was not found. When you call to method, if you get the found value you stop the loop and return the found value too.
Option B (Ugly): When found, thow an Exception (better if it is your own implementation of it). Don't forget to catch it.
Option C (Uglier): The same with global (static) variables.
UPDATE 1:
It looks like your method should run ok now, can you check (System.out.println) if your value is ever found?
In a more personal opinion, I would find
protected boolean DFS(String anaphorKey) {
for(Tree<T> child : leafs) {
if(child.head.equals(anaphorKey))
return true;
if(child.DFS(anaphorKey)) // No need to store value. No need to check == true (it is implicit)
return true; // If we are in this line the value was found, always return true
System.out.println(child.head);
//System.out.println("anaphorKey: "+anaphorKey);
}
return false; // If the method did not exit previously it was because the value was not found, so in this line always return false
}
more readable (but it should work exactly as your implementation)

What is the fastest way to compare two sets in Java?

I am trying to optimize a piece of code which compares elements of list.
Eg.
public void compare(Set<Record> firstSet, Set<Record> secondSet){
for(Record firstRecord : firstSet){
for(Record secondRecord : secondSet){
// comparing logic
}
}
}
Please take into account that the number of records in sets will be high.
Thanks
Shekhar

firstSet.equals(secondSet)
It really depends on what you want to do in the comparison logic... ie what happens if you find an element in one set not in the other? Your method has a void return type so I assume you'll do the necessary work in this method.
More fine-grained control if you need it:
if (!firstSet.containsAll(secondSet)) {
// do something if needs be
}
if (!secondSet.containsAll(firstSet)) {
// do something if needs be
}
If you need to get the elements that are in one set and not the other.
EDIT: set.removeAll(otherSet) returns a boolean, not a set. To use removeAll(), you'll have to copy the set then use it.
Set one = new HashSet<>(firstSet);
Set two = new HashSet<>(secondSet);
one.removeAll(secondSet);
two.removeAll(firstSet);
If the contents of one and two are both empty, then you know that the two sets were equal. If not, then you've got the elements that made the sets unequal.
You mentioned that the number of records might be high. If the underlying implementation is a HashSet then the fetching of each record is done in O(1) time, so you can't really get much better than that. TreeSet is O(log n).

If you simply want to know if the sets are equal, the equals method on AbstractSet is implemented roughly as below:
public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof Set))
return false;
Collection c = (Collection) o;
if (c.size() != size())
return false;
return containsAll(c);
}
Note how it optimizes the common cases where:
the two objects are the same
the other object is not a set at all, and
the two sets' sizes are different.
After that, containsAll(...) will return false as soon as it finds an element in the other set that is not also in this set. But if all elements are present in both sets, it will need to test all of them.
The worst case performance therefore occurs when the two sets are equal but not the same objects. That cost is typically O(N) or O(NlogN) depending on the implementation of this.containsAll(c).
And you get close-to-worst case performance if the sets are large and only differ in a tiny percentage of the elements.
UPDATE
If you are willing to invest time in a custom set implementation, there is an approach that can improve the "almost the same" case.
The idea is that you need to pre-calculate and cache a hash for the entire set so that you could get the set's current hashcode value in O(1). Then you can compare the hashcode for the two sets as an acceleration.
How could you implement a hashcode like that? Well if the set hashcode was:
zero for an empty set, and
the XOR of all of the element hashcodes for a non-empty set,
then you could cheaply update the set's cached hashcode each time you added or removed an element. In both cases, you simply XOR the element's hashcode with the current set hashcode.
Of course, this assumes that element hashcodes are stable while the elements are members of sets. It also assumes that the element classes hashcode function gives a good spread. That is because when the two set hashcodes are the same you still have to fall back to the O(N) comparison of all elements.
You could take this idea a bit further ... at least in theory.
WARNING - This is highly speculative. A "thought experiment" if you like.
Suppose that your set element class has a method to return a crypto checksums for the element. Now implement the set's checksums by XORing the checksums returned for the elements.
What does this buy us?
Well, if we assume that nothing underhand is going on, the probability that any two unequal set elements have the same N-bit checksums is 2-N. And the probability 2 unequal sets have the same N-bit checksums is also 2-N. So my idea is that you can implement equals as:
public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof Set))
return false;
Collection c = (Collection) o;
if (c.size() != size())
return false;
return checksums.equals(c.checksums);
}
Under the assumptions above, this will only give you the wrong answer once in 2-N time. If you make N large enough (e.g. 512 bits) the probability of a wrong answer becomes negligible (e.g. roughly 10-150).
The downside is that computing the crypto checksums for elements is very expensive, especially as the number of bits increases. So you really need an effective mechanism for memoizing the checksums. And that could be problematic.
And the other downside is that a non-zero probability of error may be unacceptable no matter how small the probability is. (But if that is the case ... how do you deal with the case where a cosmic ray flips a critical bit? Or if it simultaneously flips the same bit in two instances of a redundant system?)

There is a method in Guava Sets which can help here:
public static <E> boolean equals(Set<? extends E> set1, Set<? extends E> set2){
return Sets.symmetricDifference(set1,set2).isEmpty();
}

There's an O(N) solution for very specific cases where:
the sets are both sorted
both sorted in the same order
The following code assumes that both sets are based on the records comparable. A similar method could be based on on a Comparator.
public class SortedSetComparitor <Foo extends Comparable<Foo>>
implements Comparator<SortedSet<Foo>> {
#Override
public int compare( SortedSet<Foo> arg0, SortedSet<Foo> arg1 ) {
Iterator<Foo> otherRecords = arg1.iterator();
for (Foo thisRecord : arg0) {
// Shorter sets sort first.
if (!otherRecords.hasNext()) return 1;
int comparison = thisRecord.compareTo(otherRecords.next());
if (comparison != 0) return comparison;
}
// Shorter sets sort first
if (otherRecords.hasNext()) return -1;
else return 0;
}
}

You have the following solution from https://www.mkyong.com/java/java-how-to-compare-two-sets/
public static boolean equals(Set<?> set1, Set<?> set2){
if(set1 == null || set2 ==null){
return false;
}
if(set1.size() != set2.size()){
return false;
}
return set1.containsAll(set2);
}
Or if you prefer to use a single return statement:
public static boolean equals(Set<?> set1, Set<?> set2){
return set1 != null
&& set2 != null
&& set1.size() == set2.size()
&& set1.containsAll(set2);
}

If you are using Guava library it's possible to do:
SetView<Record> added = Sets.difference(secondSet, firstSet);
SetView<Record> removed = Sets.difference(firstSet, secondSet);
And then make a conclusion based on these.

I would put the secondSet in a HashMap before the comparison. This way you will reduce the second list's search time to n(1). Like this:
HashMap<Integer,Record> hm = new HashMap<Integer,Record>(secondSet.size());
int i = 0;
for(Record secondRecord : secondSet){
hm.put(i,secondRecord);
i++;
}
for(Record firstRecord : firstSet){
for(int i=0; i<secondSet.size(); i++){
//use hm for comparison
}
}

public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof Set))
return false;
Set<String> a = this;
Set<String> b = o;
Set<String> thedifference_a_b = new HashSet<String>(a);
thedifference_a_b.removeAll(b);
if(thedifference_a_b.isEmpty() == false) return false;
Set<String> thedifference_b_a = new HashSet<String>(b);
thedifference_b_a.removeAll(a);
if(thedifference_b_a.isEmpty() == false) return false;
return true;
}

I think method reference with equals method can be used. We assume that the object type without a shadow of a doubt has its own comparison method. Plain and simple example is here,
Set<String> set = new HashSet<>();
set.addAll(Arrays.asList("leo","bale","hanks"));
Set<String> set2 = new HashSet<>();
set2.addAll(Arrays.asList("hanks","leo","bale"));
Predicate<Set> pred = set::equals;
boolean result = pred.test(set2);
System.out.println(result); // true