MongoDB aggregation: Count documents in a query for each array field

Here’s an example through JS code of what I’m trying to achieve:

let waiting = findSessions()  // regular query for status "WAITING"
let results = [];
for (let w of waiting) {

  // Only push it to results if the w.members[0] and TARGET_USER_ID have never matched before.
  // Meaning, in the "session" collection, there are no documents that have these 2 IDs in the members field
  if (!hasMatchedBefore(w.members[0], "TARGET_USER_ID")) {
    results.push(w);
  }
}

IGNORE MOST OF WHAT’S ABOVE

Just realized how poorly written the old question was. Let’s start from the beginning.

Consider the collection "sessions":

[
    {
      status: "WAITING",
      type: "TEXT",
      members: [
        "adam"
      ]
    },
    {
      status: "WAITING",
      type: "TEXT",
      members: [
        "john"
      ]
    },
    {
      status: "WAITING",
      type: "VOICE",
      members: [
        "alex"
      ]
    },
    {
      status: "ENDED",
      type: "VOICE",
      members: [
        "adam",
        "alex"
      ]
    },
    {
      status: "TIMEOUT",
      type: "TEXT",
      members: [
        "adam",
        "TARGET"
      ]
    }
]

I’m making a match-matching system. Let’s say "TARGET" wants to match. I’m trying to write a MongoDB aggregation that does the following.

1. Find all documents with query { type: "TEXT", status: "WAITING" }
2. Iterate through each document: check if members[0] and TARGET have ever matched before
3. If members[0] and TARGET have matched before (check entire collection, any type/status), then it will not be included in the final array

It should return the following:

[
    {
      status: "WAITING",
      type: "TEXT",
      members: [
        "john"
      ]
    },
]

Notice how there were 3 "WAITING" rooms in the collection. But TARGET had already matched with adam. And TARGET cannot match with alex, because alex is in a "VOICE" session. So in this case, john would be the only appropriate match.