Stock Price Fluctuation

Published: Sep 7, 2022

Medium Heap Hash Table Design Data Stream Ordered Set

Introduction

This problem is a different type of stocks. Stock prices are coming in one by one from data stream. It means this is an online algorithm type. Additionally, this is a design problem. A class template is given with empty constructor and methods.

Problem Description

You are given a stream of `record`s 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 the price of the stock at the given timestamp.
• `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.

Constraints:

`1 <= timestamp, price <= 10**9`
At most `10**5` calls will be made in total to `update`, `current`, `maximum`, and `minimum`. `current`, `maximum`, and `minimum` will be called only after `update` has been called at least once.

https://leetcode.com/problems/stock-price-fluctuation/

Examples

``````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.
``````

Analysis

If the problem is the type of online algorithm and asks maximum and/or minimum, the heap data structure should be used. The problem asks both maximum and minimum, we need min and max heaps. To support current value reference, we should keep values in a hash map.

Each method should focus on only necessary operations. Heap updates will be done when maximum or minimum methods are called. In max/min heaps, old data are saved. Those will be eliminated when heap pop result mismatches values in hash map.

Solution

``````class StockPrice:

def __init__(self):
self.curId = -1
self.d = {}  # id: price
self.maxheap = [] # (price, id)
self.minheap = [] # (price, id)

def update(self, timestamp: int, price: int) -> None:
self.curId = max(self.curId, timestamp)
self.d[timestamp] = price
heapq.heappush(self.maxheap, (-price, timestamp))
heapq.heappush(self.minheap, (price, timestamp))

def current(self) -> int:
return self.d[self.curId]

def maximum(self) -> int:
p, id = self.maxheap[0]
while self.maxheap and self.d[id] != -p:
heapq.heappop(self.maxheap)
p, id = self.maxheap[0]
return -p

def minimum(self) -> int:
p, id = self.minheap[0]
while self.minheap and self.d[id] != p:
heapq.heappop(self.minheap)
p, id = self.minheap[0]
return p
``````

Complexities

• Time: `O(nlog(n))`
• Space: `O(n)`