Challenge 2: Broadcast a "Left" Presence Event — Possible Solution
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function leaveRoom(socket, roomName) {
  const members = rooms.get(roomName);
  if (!members) return;

  members.delete(socket);
  socket.rooms.delete(roomName);

  // Broadcast to whoever is STILL in the room after this socket has
  // already been removed from it.
  sendToRoom(roomName, JSON.stringify({
    type: "presence", event: "left", userId: socket.userId
  }));
}

WHY THIS WORKS
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- members.delete(socket) and socket.rooms.delete(roomName) both happen
  BEFORE sendToRoom() is called — this ordering matters: sendToRoom()
  reads whoever is currently in the room's Set, and the departing socket
  should not be one of the recipients of its own "left" announcement.
- Reusing sendToRoom() (rather than writing a separate broadcast loop
  here) means this "left" event gets the exact same readyState-checked,
  safe-broadcast behavior as every other message sent to the room — one
  broadcast implementation, used consistently everywhere a room needs to
  be notified of something.
- The message shape — { type: "presence", event: "left", userId } —
  deliberately mirrors the "joined" event's shape from the chapter
  (only the event field differs). This consistency matters on the
  client side: a single presence-handling function can switch on
  event ("joined" vs "left") rather than needing entirely separate
  message formats for the two cases.
- If members is undefined (the room never existed, or was already fully
  cleaned up), the function returns early — there's nothing to remove
  and no one left in a nonexistent room to notify, so this guards against
  calling .delete() on undefined or broadcasting to a room with no Set.
- Note that by the time this runs, socket.userId still holds a valid
  value even though the socket has just been removed from the room —
  deleting from a Set doesn't erase properties on the socket object
  itself, so the announcement can still correctly identify who left.