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.
Constraints:
- 1 <= n <= 1000
- 1 <= id <= n
- value.length == 5
- value consists only of lowercase letters.
- Each call to insert will have a unique id.
- Exactly n calls will be made to insert.
Solution in python:
class OrderedStream:
def __init__(self, n: int):
self.items = [None for i in range(n)]
self.ptr = 0
def insert(self, idKey: int, value: str) -> List[str]:
self.items[idKey-1] = value
i = self.ptr
while i < len(self.items) and self.items[i] != None:
i += 1
result = self.items[self.ptr:i]
self.ptr = i
return result
# Your OrderedStream object will be instantiated and called as such:
# obj = OrderedStream(n)
# param_1 = obj.insert(idKey,value)
留言