Friend groups by matching last k movies

A streaming service stores each customer's full viewing history, oldest title first. Two customers are friends when the set of their last k watched titles is the same. Order inside that window does not matter, and a title watched twice in the window counts once.

Implement:

def friend_groups(customer_ids: list[str], movies: list[list[str]], k: int) -> list[list[str]]

• customer_ids holds distinct IDs; movies[i] is the history of customer_ids[i].
• Return every friend group with at least two customers. Sort the IDs inside each group, and sort the groups by their first ID.
• A customer who has watched fewer than k titles has no full window and belongs to no group.

Constraints: 1 <= k <= 100, up to 10^5 customers, histories up to 10^4 titles each.

Example 1

customer_ids = ["u1", "u2", "u3", "u4"]
movies = [
  ["Heat", "Alien", "Up", "Jaws"],
  ["Jaws", "Up", "Alien"],
  ["Alien", "Jaws", "Up"],
  ["Jaws", "Heat", "Up"],
]
k = 3  ->  [["u1", "u2", "u3"]]

u1, u2 and u3 all end with the set {Alien, Jaws, Up}. u4 ends with {Jaws, Heat, Up}.

Example 2

Same input with k = 1 returns [["u3", "u4"]]: both last watched Up.

Be ready for follow-ups: friends who share at least m of the last k titles, histories that grow as events stream in, and data too large for one machine.