List<Batch> vAllBatchList = getAllBatchCollection().toList(); //Has 700k records
List<Batch> vKeepableBatchCollection = getKeepableBatchCollection(pDaysKeepHistory).toList(); //has 600k records
vAllBatchList.removeAll(vKeepableBatchCollection);
In the above 3rd line removeAll method is taking too much time to finish. How to optimize the removeAll method here?
If you convert the List of element to remove to a Set, it should be faster:
vAllBatchList.removeAll(new HashSet<>(vKeepableBatchCollection));
This is assuming Batch class overrides hashCode and equals properly.
Explanation: removeAll for ArrayList (I'm assuming your vAllBatchList List is an ArrayList) iterates over all the elements of the List on which it is called, and checks if the passed Collection contains them. If the passed Collection is a Set, contains will take expected constant time (O(1)), while if the Collection is a List, it will take linear time (O(n)).
Of course, if you can directly generate a Set of the elements of vKeepableBatchCollection instead of first creating a List and then converting it to a Set, it would be even better.