花花酱 LeetCode 1656. Design an Ordered Stream

There are n (id, value) pairs, where id is an integer between 1 and n and value is a string. No two pairs have the same id.

Design a stream that takes the n pairs in an arbitrary order, and returns the values over several calls in increasing order of their ids.

Implement the OrderedStream class:

• OrderedStream(int n) Constructs the stream to take n values and sets a current ptr to 1.
• String[] insert(int id, String value) Stores the new (id, value) pair in the stream. After storing the pair:
  • If the stream has stored a pair with id = ptr, then find the longest contiguous incrementing sequence of ids starting with id = ptr and return a list of the values associated with those ids in order. Then, update ptr to the last id + 1.
  • Otherwise, return an empty list.

Example:

Input
["OrderedStream", "insert", "insert", "insert", "insert", "insert"]
[[5], [3, "ccccc"], [1, "aaaaa"], [2, "bbbbb"], [5, "eeeee"], [4, "ddddd"]]
Output
[null, [], ["aaaaa"], ["bbbbb", "ccccc"], [], ["ddddd", "eeeee"]]
Explanation
OrderedStream os= new OrderedStream(5);
os.insert(3, "ccccc"); // Inserts (3, "ccccc"), returns [].
os.insert(1, "aaaaa"); // Inserts (1, "aaaaa"), returns ["aaaaa"].
os.insert(2, "bbbbb"); // Inserts (2, "bbbbb"), returns ["bbbbb", "ccccc"].
os.insert(5, "eeeee"); // Inserts (5, "eeeee"), returns [].
os.insert(4, "ddddd"); // Inserts (4, "ddddd"), returns ["ddddd", "eeeee"].

Solution: Straight Forward

Time complexity: O(n) in total
Space complexity: O(n)