group array of objects and get the count in mongodb

I am having an array of objects:

[
  {
    medicationDetails: {
	  category: 'Asthma',
	  subCategory: '',
	  providedBy: 'Doctor'
    }
  },
  {
    medicationDetails: {
	  category: 'Diabetes',
	  subCategory: 'Oral',
	  providedBy: 'Nurse'
    }
  },
  {
    medicationDetails: {
	  category: 'Asthma',
	  subCategory: ''
	  providedBy: 'Doctor'
    }
  },
  {
    medicationDetails: {
	  category: 'Diabetes',
	  subCategory: 'Insulin',
	  providedBy: 'Doctor'
    }
  }
]

Some medication will have sub-category and some doesn't.

Need to group based on 'category', 'sub-category' and 'providedBy' and get the count like this:

[
  {
    medicationDetails: [
     {
      name: 'Asthma',
	  providedByDoctorCount: 2,
	  providedByNurseCount: 0
     },
     {
      name: 'Oral',
      providedByDoctorCount: 0,
	  providedByNurseCount: 1
     },
     {
      name: 'Insulin',
      providedByDoctorCount: 1,
	  providedByNurseCount: 0
     }
    ]
  }
]

I was tried with $group and $count but not getting the expected result.
Need some valuable help.

somewhat similar to your previous question.
you can conditionally sum based on the value of the field when grouping

1. Create a key field based on category and subCategory values
2. Group by the key and add to providedByDoctor or providedByNure conditionally based on the value of providedBy
3. Group into a single medicationDetails array

db.collection.aggregate([
  { 
   $project: {
      providedBy: "$medicationDetails.providedBy",
      key: { $cond: [ { $eq: [ "$medicationDetails.subCategory", "" ] }, "$medicationDetails.category", "$medicationDetails.subCategory" ] }
    } 
  },
  {
    $group: { 
      _id: "$key",
      providedByDoctor: { $sum: { $cond: [ { $eq: [ "$providedBy", "Doctor" ] }, 1, 0 ] } },
      providedByNurse: { $sum: { $cond: [ { $eq: [ "$providedBy", "Nurse" ] }, 1, 0 ] } } }
  },
  { $group: { _id: null, medicationDetails: { $push: { name: "$_id", providedByDoctorCount: "$providedByDoctor", providedByNurseCount: "$providedByNurse" } } } },
  { $unset: "_id" }
])

playground