Leet code problem 1656: – Design an Ordered Stream

There is a stream of n (idKey, value) pairs arriving in an arbitrary order, where idKey is an integer between 1 and n and value is a string. No two pairs have the same id.

Design a stream that returns the values in increasing order of their IDs by returning a chunk (list) of values after each insertion. The concatenation of all the chunks should result in a list of the sorted values.

Implement the OrderedStream class:

OrderedStream(int n) Constructs the stream to take n values.
String[] insert(int idKey, String value) Inserts the pair (idKey, value) into the stream, then returns the largest possible chunk of currently inserted values that appear next in the order.

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
// Note that the values ordered by ID is [“aaaaa”, “bbbbb”, “ccccc”, “ddddd”, “eeeee”].
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”].
// Concatentating all the chunks returned:
// [] + [“aaaaa”] + [“bbbbb”, “ccccc”] + [] + [“ddddd”, “eeeee”] = [“aaaaa”, “bbbbb”, “ccccc”, “ddddd”, “eeeee”]
// The resulting order is the same as the order above.

Flow Diagram: –

Solution: –

class OrderedStream {
    int ptr;
    String[] res;
    
    public OrderedStream(int n) {
        ptr = 0;
        res = new String[n];
    }
    
    public List<String> insert(int id, String value) {
        List<String> list = new ArrayList<>();
        
        res[id - 1] = value;
        while (ptr < res.length && res[ptr] != null) {
            list.add(res[ptr]);
            ptr++;
        }
        
        return list;
    }
}