hashmap - Intersection of a Set with a twist Java -

so got following question in technical interview , thought pretty interesting.

given 2 arrays: = {1,2,2,4,5} , b = {1,2,6,} write code performs following 2 operations: a-b {2,4,5} , b-a {6}. meaning, find common element in other array , remove original. if b {1,2}, b-a {} since contains elements of b , b has no unique elements.

i solved question in following way, , hoping see if had suggestions on how possibly make better.

public static void main(string args[]) { map hashtable = new hashmap();  int a[] = {1,1,2,4,5,6,6}; int b[] = {1,2,6};  arraylist ba = new arraylist(); arraylist ab = new arraylist();  int[] occurances = new int[a.length]; //int occurances = 0; (int = 0; < a.length; i++) {     occurances[a[i]]++;     hashtable.put(a[i], occurances[a[i]]);  } //system.out.println(arrays.tostring(occurances)); system.out.println(hashtable); //find ba (int = 0; < b.length; i++) {     if(hashtable.containskey(b[i])) {         occurances[b[i]]--;         hashtable.put(b[i], occurances[b[i]]);     } else ba.add(b[i]);   } for(int = 0; <a.length; i++) {     if(hashtable.containskey(a[i]) && occurances[a[i]] != 0) {         ab.add(a[i]);         occurances[a[i]]--;         hashtable.put(a[i], occurances[a[i]]);      }  }  system.out.println("ab = " + ab); system.out.println("ba =" + ba);   }   } 

****edit***** mistakenly called arrays sets when posed question. since arrays can have duplicate elements, definition not sets. sorry confusion.

you can use set have union , intersection functions.

integer a[] = {1, 2, 2, 4, 5}; integer b[] = {1, 2, 6};  public void test() {     // grow 2 sets arrays.     set<integer> sa = arrays.stream(a)             .collect(collectors.tocollection(treeset::new));     set<integer> sb = arrays.stream(b)             .collect(collectors.tocollection(treeset::new));     // make new 1 don't damage sa or sb.     set<integer> sc = new hashset<>(sa);     sc.removeall(sb);     system.out.println(sa + " - " + sb + " = " + sc);     set<integer> sd = new hashset<>(sb);     sd.removeall(sa);     system.out.println(sb + " - " + sa + " = " + sd); } 

prints

[1, 2, 4, 5] - [1, 2, 6] = [4, 5] [1, 2, 6] - [1, 2, 4, 5] = [6]