You are given a list of users' movie-watching histories. Each user’s history is an ordered list of movie IDs in the order they were watched (earliest to latest). Assume: - `histories` is a dictionary/map from `user_id` (string or integer) to a list of `movie_id` (string or integer). - Each list has at least `K` movies. **Task** Write an algorithm to group users such that, within each group, all users share the exact same set of the last `K` movies they watched (ignoring order within those last `K` movies). Two users belong to the same group if the following holds: - Let `S(u)` be the set of the last `K` movies in user `u`'s history. - Users `u1` and `u2` are in the same group if `S(u1) = S(u2)`. **Input** - An integer `K` (1 ≤ K ≤ 10^3). - A map/dictionary `histories: user_id → [movie_id1, movie_id2, ..., movie_idN]` where N ≥ K for each user. **Output** - Return a list of groups, where each group is a list of `user_id`s. Only groups of size ≥ 2 (i.e., at least two users with the same last-K set) need to be returned. **Example** Suppose K = 3 and: - user A: [m1, m2, m3, m4, m5] - user B: [m7, m3, m4, m5] - user C: [m2, m3, m4, m6] Then: - Last 3 for A: {m3, m4, m5} - Last 3 for B: {m3, m4, m5} - Last 3 for C: {m3, m4, m6} The output should group A and B together, and C can be omitted or appear alone depending on your choice, but only multi-user groups are required in the output. Describe the algorithm and its time and space complexity.

def solution(K, histories):
    groups = {}

    for user_id, watched in histories.items():
        key = frozenset(watched[-K:])
        if key not in groups:
            groups[key] = []
        groups[key].append(user_id)

    result = []
    for users in groups.values():
        if len(users) >= 2:
            users.sort(key=lambda x: str(x))
            result.append(users)

    result.sort(key=lambda group: tuple(str(x) for x in group))
    return result