Single mongo operation to remove subdocuments based on certain condition in spring data mongodb

I am new to mongodb. I have a document structure like this in mongo collection

{
  orgId: "",
  location: "",
  companyId: "",
  employees: [
    { 
      employeeId: "",
      status: "",
      startDate: ""
    },
    {
      employeeId: "",
      status: "",
      promotionDate: ""
    }
  ]
}

And wanted to write a query to find all the documents having location="US", statuses="{L1, L2}" who all promoted before today and remove all the subdocuments matching criteria.

since I don't have IDs here to pull out employees I thought of using aggregation but it's not working.

        
Date currentTime = Date.from(Instant.now());
MatchOperation locationMatch = Aggregation.match(Criteria.where("location").in(locations));
        UnwindOperation unwindOperation = Aggregation.unwind("employees");
        MatchOperation logMatch = Aggregation.match(Criteria.where("employees.statuses").in(statuses));
        MatchOperation expiryMatch = Aggregation.match(Criteria.where("employees. promotionDate").lt(currentTime));

        Aggregation aggregationPipeline = Aggregation.newAggregation(
                locationMatch,
                unwindOperation,
                logMatch,
                expiryMatch);

        List<Employees> employees =
                mongoTemplate.aggregate(aggregationPipeline, Company.class, employees.class)
                        .getMappedResults();


        List<String> employeeList = employees.stream()
                .map(emp -> emp.employeeId())
                .collect(Collectors.toList());

        Update update = new Update().pull("list", Query.query(Criteria.where("value").in(employeeList)));

        return mongoTemplate.findAndModify(new Query(), update,
                new FindAndModifyOptions().returnNew(true), Company.class);

Is there anyway I can achieve this result in a single mongo operation? or perhaps how would I delete all the subdocuments matching criteria?