Aquí hay una solución que funciona explotando $lookup en la misma colección (piense en "auto-JOIN"):

 var r = [ {_id: 0, aa: [ 10, 11, 12 ] } ,{_id: 1, aa: [ 10, 11, 12 ] } ,{_id: 2, aa: [ 20, 21, 22 ] } // 21 is on watch list... ,{_id: 3, aa: [ 21, 20, 12 ] } // this one too and 21 is in different position ,{_id: 4, aa: [ 10, 22, 12 ] } ,{_id: 5, aa: [ 10, 22, 23 ] } // this one too... ,{_id: 6, aa: [ 10, 22, 21, 23 ] } // this one has BOTH 21 and 23 ,{_id: 10, X:10} ,{_id: 11, X:11} ,{_id: 12, X:12} ,{_id: 20, X:20} ,{_id: 21, X:21} ,{_id: 22, X:22} ,{_id: 23, X:23} ]; db.foo.insert(r); // Here is the whole thing: db.foo.aggregate([ ]); // Delete _id 21 and 23: db.foo.remove({_id: 21}); db.foo.remove({_id: 23}); // Double check: c = db.foo.aggregate([ ]); // Where does id 21 and/or 23 NOT exist anymore? Note we don't ask for 21 or 23. // We just know we expect a query to return docs that indicate 21 and/or 23 // are no longer there: c = db.foo.aggregate([ // NOTE! By using localField:'aa', we are asking for EACH element in the // array to be used as a value to match to _id (in the same collection): {$lookup: {from: 'foo', localField: 'aa', foreignField: '_id', as: 'X'}}, // Exploit "make a list of scalars from array of objects" notation by taking // input array $X and taking the _id field out: {$project: {X: {$setDifference: ["$aa", "$X._id"] }} }, // Keep those that match -- and protect against against empty sets // with $ifNull to turn a null into an array of len 0: {$match: {$expr: {$gt:[{$size: {$ifNull:['$X',[]]}}, 0]}} } ]); { "_id" : 2, "X" : [ 21 ] } { "_id" : 3, "X" : [ 21 ] } { "_id" : 5, "X" : [ 23 ] } { "_id" : 6, "X" : [ 21, 23 ] }