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The fastest way to do a subtraction collection

I have two sets. Set b- a subset Set a. They are both very large Sets. I want to subtract b from what is best done for this general operation? I have written a lot of codes like this and I do not find it effective. what is your idea

pseudo code: (this is not a Java API).

for(int i = 0 ; i < a.size(); i++) {
          for (int j=0 ; j < b.size() ;j++) {
              // do comparison , if found equals ,remove from a
              break;
          }
 }

And I want to find an algorithm that not only applies to Sets, but also works for Array.

EDIT: the set here is not a JAVA API, it is a data structure. so I don’t care if the Java API has a removeAll () method, I want to find a common solution for this problem, I am faced with a lot of problems like this when I use Javascript and Actionscript.

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int bIndex = 0;
for(int i = 0 ; i < a.size(); i++) {
          while (a[i] < b[bIndex]) {bIndex++;}
          if (a[i] == b[bIndex]) {markForRemoval(a[i]);} // I mark this only for removal, as the actual removal would make your index incorrect
}

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:

for each (var key:* in set2) {//a simple for-in loop will also do the trick, since keys and values are equal, but for-each-in loops perform faster
    delete set1[key];
}

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:

for (var key in set2) {
    delete set1[key];
}
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foreach b in B
    remove b from A

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removeAll Set?

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// A is some kind of array, O(1) iteration
// B is a hash containing elements to remove, O(1) contains(elt)
List<T> removeAll(List<T> A, Set<T> B) {
  List<T> result; // empty, could preallocate at |A|
  for (elt : A) { // for each 'elt' belonging to A, hence O(|A|)
    if (! B.contains(elt) ) { // O(1) thanks to hash
      C.add(elt) ; // ensure this is O(1) with preallocation or linked list
    }
  }
  return result;
}

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