Find a direct room with one query instead of checking every one

Opening a direct room looked for an existing one by loading every direct
room on the account and comparing its member ids in Ruby, two queries per
room, on each click. The cost grows with the account: about 0.3 s at 1,000
direct rooms and 3 s at 10,000. Asking SQL for the room among the first
user's memberships whose member set is exactly the given users finds the
same room in one query, however many direct rooms there are.

Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_01JVFo3Lt9T8M5NR7KxVvsZ2

app/models/rooms/direct.rb

@@ -7,12 +7,15 @@ class Rooms::Direct < Room
     end
 
     private
-      # FIXME: Find a more performant algorithm that won't be a problem on accounts with 10K+ direct rooms,
-      # which could be to store the membership id list as a hash on the room, and use that for lookup.
+      # Among the first user's rooms, the one whose members are exactly these users: as many
+      # memberships as users, and all of them theirs.
       def find_for(users)
-        all.joins(:users).detect do |room|
-          Set.new(room.user_ids) == Set.new(users.pluck(:id))
-        end
+        user_ids = users.pluck(:id).uniq
+
+        where(id: Membership.where(user_id: user_ids.first).select(:room_id))
+          .joins(:memberships).group(:id)
+          .having("COUNT(*) = :size AND COUNT(CASE WHEN memberships.user_id IN (:user_ids) THEN 1 END) = :size", size: user_ids.size, user_ids: user_ids)
+          .first
       end
   end
 

test/models/rooms/direct_test.rb

@@ -18,4 +18,33 @@ class Rooms::DirectTest < ActiveSupport::TestCase
     room = Rooms::Direct.find_or_create_for([ users(:david), users(:kevin) ])
     assert room.memberships.all? { |m| m.involved_in_everything? }
   end
+
+  test "find the room with exactly the same users" do
+    Current.user = users(:david)
+    group = Rooms::Direct.find_or_create_for([ users(:david), users(:kevin), users(:jason) ])
+
+    assert_equal rooms(:david_and_kevin), Rooms::Direct.find_or_create_for([ users(:kevin), users(:david) ])
+    assert_equal group, Rooms::Direct.find_or_create_for(User.where(id: [ users(:jason), users(:kevin), users(:david) ]))
+    assert_equal [ users(:david), users(:jz) ].to_set, Rooms::Direct.find_or_create_for([ users(:david), users(:jz) ]).users.to_set
+    assert_equal [ users(:jason), users(:kevin) ].to_set, Rooms::Direct.find_or_create_for([ users(:jason), users(:kevin) ]).users.to_set
+  end
+
+  test "finding a room takes the same queries however many direct rooms the account has" do
+    Current.user = users(:david)
+    Rooms::Direct.create_for({}, users: [ users(:david), users(:jz) ])
+    before = queries_during { Rooms::Direct.find_or_create_for(User.where(id: [ users(:david), users(:jz) ])) }
+
+    5.times { |i| Rooms::Direct.create_for({}, users: [ users(:jason), User.create!(name: "Guest #{i}", email_address: "guest#{i}@example.com", password: "secret123456") ]) }
+    Rooms::Direct.create_for({}, users: [ users(:kevin), users(:jz) ])
+
+    assert_equal before, queries_during { Rooms::Direct.find_or_create_for(User.where(id: [ users(:kevin), users(:jz) ])) }
+  end
+
+  private
+    def queries_during(&block)
+      count = 0
+      counter = ->(*, payload) { count += 1 unless payload[:name] == "SCHEMA" || payload[:cached] }
+      ActiveSupport::Notifications.subscribed(counter, "sql.active_record", &block)
+      count
+    end
 end