Movie History Friends I / II

Two related prompts over per-customer watch histories: Part I pairs customers whose last k movies match exactly (same order); Part II pairs customers sharing at least m of their last k watches in any order. Return pairs with the smaller ID first, no duplicates. Part I — Exact Last-k Match Problem Challenge: Movie History Friends Every customer has a list of movies they have watched. The list keeps the specific order in which they viewed them. We define two customers as "friends" if their last k movies are an exact match. This means the movies and their order must be identical. You are given a map that links customer IDs to their movie history, and a number k. Your task is to find and return all pairs of customers who are friends. Important Rules Not enough movies: If a customer has watched fewer than k movies, they cannot be friends with anyone. ID Order: You can return the list of pairs in any order. However, inside each pair, the smaller ID must come first (for example, use [1, 2], not [2, 1]). No Duplicates: Each pair of friends should appear only once in your output. Sample Cases Case 1: Input: history = {1: ["M1", "M2", "M3", "M4", "M5"], 2: ["X", "Y", "M3", "M4", "M5"], 3: ["A", "B", "M3", "M4", "M5"]}, k = 3 Output: [[1, 2], [1, 3], [2, 3]] Why? All three customers watched the same last 3 movies: ["M3", "M4", "M5"]. The order is the same for everyone, so every customer is friends with every other customer. Case 2: Input: history = {1: ["A", "B", "C"], 2: ["X", "B", "C"], 3: ["A", "B", "C"]}, k = 3 Output: [[1, 3]] Why? Only customers 1 and 3 have the exact same last 3 movies (["A", "B", "C"]). Customer 2 is different because their first movie does not match the others. Case 3: Input: history = {1: ["A", "B"], 2: ["C", "D", "E"]}, k = 3 Output: [] Why? Customer 1 has only watched 2 movies. Since k is 3, they do not have enough history to match anyone. No pairs are possible. Input Limits 1 <= number of customers <= 1000 1 <= k <= 100 0 <= history[i].length <= 1000 Movie IDs are strings Part II — At Least m of Last k (Unordered) The Challenge We have a record of movies watched by different customers. The list for each customer shows the movies in the exact order they watched them. In this problem, two customers are called "friends" if they have watched at least m of the same movies within their most recent k watches. These common movies do not need to be in the same order or the same spot in the list. You are given a map connecting Customer IDs to their movie history, along with the numbers k and m (where m is less than or equal to k). Your task is to return all pairs of Customer IDs that qualify as friends. Important Rules Length Check: If a customer has watched fewer than k movies total, they cannot be friends with anyone. Order Does Not Matter: The matching movies can be in any position within the last k entries. Output Format: You can list the friend pairs in any order. However, inside each pair, you must place the smaller ID first (e.g., [1, 2], not [2, 1]). No Duplicates: Each pair of friends should be listed only once. Walkthrough Examples Case 1: Input: history = {1: ["A", "B", "C", "D"], 2: ["X", "D", "B", "A"], 3: ["P", "Q", "R", "S"]}, k = 3, m = 2 Output: [[1, 2]] Breakdown: Customer 1: The last 3 movies are ["B", "C", "D"]. Customer 2: The last 3 movies are ["D", "B", "A"]. Result: They both watched "B" and "D". That is 2 common movies. Since m is 2, this is enough to be friends. Customer 3: Their movies are completely different, so they have no friends here. Case 2: Input: history = {1: ["A", "B", "C"], 2: ["C", "B", "A"], 3: ["A", "B", "C"]}, k = 3, m = 3 Output: [[1, 2], [1, 3], [2, 3]] Breakdown: All customers watched "A", "B", and "C" in their last 3 slots, just in different orders. Because they all share 3 movies, every possible pair is a friend pair. Case 3: Input: history = {1: ["A", "B", "C"], 2: ["D", "E", "F"]}, k = 3, m = 1 Output: [] Breakdown: There are no common movies between the two customers. Therefore, the result is an empty list. Data Constraints 1 <= number of customers <= 1000 1 <= m <= k <= 100 0 <= history[i].length <= 1000 Movie IDs are strings