Delete documents from collection based on condition - mongodb

Question

I want to delete the documents from a collection(Collection1) which are not matching with other collection(Collection2).
collection1 document - {_id: <autogen>, items: [{key:'key', ...}}]
collection2 document - {_id: 'key', action: {key1:'key1', ...}}, {_id: 'non_matching_key', action: {key1:'key1', ...}}
Delete all the documents from collection2, where items.key in collection1 is not matching with _id in collection2. with reference to the above example, a document with key-value 'non_matching_key' should be delete from collection2. There would be similar documents in collection2 like the one with _id value 'non_matching_key'.
The approach I thought was for mark and sweep.
I will add the column in collection2 documents for matching ids(in collection2 with items.key in collection1). This is mark step
Delete all the documents from collection2 where newly added column do not exists. This is sweep step.
Could you please advise if there is a better way of doing this?
Thanks,

Answer 1

Not fully clear how your documents look like and what are the exact conditions, but you could do it like this:
var ids = db.collection1.find({}, { "item.key": 1 }).toArray();
db.collection2.deleteMany({ _id: { $nin: ids } });

Answer 2

So basically, you want to iterate over your collection 2 and get all the ids that are not in collection 1 and delete those or vice versa.
db.collection2.find({}).forEach((doc) => db.collection1.deleteOne({_id: {$ne: doc._id}}))
or
let idsToDelete = db.collection2.find({}).distinct('_id')
let deleteResponse = db.collection1.deleteMany({_id: {$nin: idsToDelete}})
SWAP the collection name in the case of the other way.
NOTE: The code is just to give an overview and is not tested.