2034. Stock Price Fluctuation
Description
You are given a stream of records about a particular stock. Each record contains a timestamp and the corresponding price of the stock at that timestamp.
Unfortunately due to the volatile nature of the stock market, the records do not come in order. Even worse, some records may be incorrect. Another record with the same timestamp may appear later in the stream correcting the price of the previous wrong record.
Design an algorithm that:
- Updates the price of the stock at a particular timestamp, correcting the price from any previous records at the timestamp.
- Finds the latest price of the stock based on the current records. The latest price is the price at the latest timestamp recorded.
- Finds the maximum price the stock has been based on the current records.
- Finds the minimum price the stock has been based on the current records.
Implement the StockPrice
class:
StockPrice()
Initializes the object with no price records.void update(int timestamp, int price)
Updates theprice
of the stock at the giventimestamp
.int current()
Returns the latest price of the stock.int maximum()
Returns the maximum price of the stock.int minimum()
Returns the minimum price of the stock.
Example 1:
Input ["StockPrice", "update", "update", "current", "maximum", "update", "maximum", "update", "minimum"] [[], [1, 10], [2, 5], [], [], [1, 3], [], [4, 2], []] Output [null, null, null, 5, 10, null, 5, null, 2] Explanation StockPrice stockPrice = new StockPrice(); stockPrice.update(1, 10); // Timestamps are [1] with corresponding prices [10]. stockPrice.update(2, 5); // Timestamps are [1,2] with corresponding prices [10,5]. stockPrice.current(); // return 5, the latest timestamp is 2 with the price being 5. stockPrice.maximum(); // return 10, the maximum price is 10 at timestamp 1. stockPrice.update(1, 3); // The previous timestamp 1 had the wrong price, so it is updated to 3. // Timestamps are [1,2] with corresponding prices [3,5]. stockPrice.maximum(); // return 5, the maximum price is 5 after the correction. stockPrice.update(4, 2); // Timestamps are [1,2,4] with corresponding prices [3,5,2]. stockPrice.minimum(); // return 2, the minimum price is 2 at timestamp 4.
Constraints:
1 <= timestamp, price <= 109
- At most
105
calls will be made in total toupdate
,current
,maximum
, andminimum
. current
,maximum
, andminimum
will be called only afterupdate
has been called at least once.
Solutions
Solution 1: Hash Table + Ordered Set
We define the following data structures or variables:
d
: a hash table that stores the timestamp and the corresponding price;ls
: an ordered set that stores all prices;last
: the timestamp of the last update.
Then, we can perform the following operations:
update(timestamp, price)
: update the price corresponding to the timestamptimestamp
toprice
. Iftimestamp
already exists, we need to first remove its corresponding price from the ordered set, and then update it toprice
. Otherwise, we directly update it toprice
. Then, we need to updatelast
tomax(last, timestamp)
. The time complexity is O(log n).current()
: return the price corresponding tolast
. The time complexity is $O(1)$.maximum()
: return the maximum value in the ordered set. The time complexity is $O(\log n)$.minimum()
: return the minimum value in the ordered set. The time complexity is $O(\log n)$.
The space complexity is $O(n)$, where $n$ is the number of update
operations.
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