Maximum profit of stocks
On-Site Question 1 - SOLUTION
Problem
You've been given a list of historical stock prices for a single day for Amazon stock. The index of the list represents the timestamp, so the element at index of 0 is the initial price of the stock, the element at index 1 is the next recorded price of the stock for that day, etc. Your task is to write a function that will return the maximum profit possible from the purchase and sale of a single share of Amazon stock on that day. Keep in mind to try to make this as efficient as possible.
For example, if you were given the list of stock prices:
prices = [12,11,15,3,10]
Then your function would return the maximum possible profit, which would be 7 (buying at 3 and selling at 10).
Requirements
Try to solve this problem with paper/pencil first without using an IDE. Also keep in mind you should be able to come up with a better solution than just brute forcing every possible sale combination
Also you can't "short" a stock, you must buy before you sell the stock.
Solution
Let's think about a few things before we start coding. One thing to think about right off the bat is that we can't just find the maximum price and the lowest price and then subtract the two, because the max could come before the min.
The brute force method would be to try every possible pair of price combinations, but this would be O(N^2), pretty bad. Also since this is an interview setting you should probably already know that there is a smarter solution.
In this case we will use a greedy algorithm approach. We will iterate through the list of stock prices while keeping track of our maximum profit.
That means for every price we will keep track of the lowest price so far and then check if we can get a better profit than our current max.
Let's see an implementation of this:
def profit(stock_prices): # Start minimum price marker at first price
min_stock_price = stock_prices[0] # Start off with a profit of zero
max_profit = 0 for price in stock_prices: # Check to set the lowest stock price so far
min_stock_price = min(min_stock_price,price) # Check the current price against our minimum for a profit
# comparison against the max_profit
comparison_profit = price - min_stock_price # Compare against our max_profit so far
max_profit = max(max_profit,comparison_profit) return max_profit
profit([10,12,14,12,13,11,8,7,6,13,23,45,11,10])
39
Currently we're finding the max profit in one pass O(n) and in constant space O(1). However, we still aren't thinking about any edge cases. For example, we need to address the following scenarios:
- Stock price always goes down
- If there's less than two stock prices in the list.
We can take care of the first scenario by returning a negative profit if the price decreases all day (that way we can know how much we lost). And the second issue can be solved with a quick len() check. Let's see the full solution:
def profit2(stock_prices): # Check length
if len(stock_prices) < 2:
raise Exception('Need at least two stock prices!') # Start minimum price marker at first price
min_stock_price = stock_prices[0] # Start off with an initial max profit
max_profit = stock_prices[1] - stock_prices[0] # Skip first index of 0
for price in stock_prices[1:]: # NOTE THE REORDERING HERE DUE TO THE NEGATIVE PROFIT TRACKING # Check the current price against our minimum for a profit
# comparison against the max_profit
comparison_profit = price - min_stock_price # Compare against our max_profit so far
max_profit = max(max_profit,comparison_profit) # Check to set the lowest stock price so far
min_stock_price = min(min_stock_price,price) return max_profit
# Exception Raised
profit2([1])
---------------------------------------------------------------------------
Exception Traceback (most recent call last)
<ipython-input-11-7bd2f0c7e63b> in <module>()
1 # Exception Raised
----> 2 profit2([1]) <ipython-input-10-e06adf3c45a7> in profit2(stock_prices)
3 # Check length
4 if len(stock_prices) < 2:
----> 5 raise Exception('Need at least two stock prices!')
6
7 # Start minimum price marker at first price Exception: Need at least two stock prices!
profit2([30,22,21,5])
-1
Great! Now we can prepare for worst case scenarios. Its important to keep edge cases in mind, especially if you are able to solve the original question fairly quickly.
Good Job!
Maximum profit of stocks的更多相关文章
- [Educational Round 59][Codeforces 1107G. Vasya and Maximum Profit]
咸鱼了好久...出来冒个泡_(:з」∠)_ 题目连接:1107G - Vasya and Maximum Profit 题目大意:给出\(n,a\)以及长度为\(n\)的数组\(c_i\)和长度为\( ...
- Yaoge’s maximum profit HDU - 5052
Yaoge’s maximum profit Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/ ...
- Maximum Profit
Maximum Profit You can obtain profits from foreign exchange margin transactions. For example, if you ...
- Codeforces 1107G Vasya and Maximum Profit 线段树最大子段和 + 单调栈
Codeforces 1107G 线段树最大子段和 + 单调栈 G. Vasya and Maximum Profit Description: Vasya got really tired of t ...
- 【leetcode】1235. Maximum Profit in Job Scheduling
题目如下: We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtai ...
- HDU5052 Yaoge’s maximum profit(LCT)
典型的LCT操作,但是维护的是一个序列最左边减最右边的最小值,所以要维护左边减右边的最小值del[0]和一个右边减左边的最小值del[1](因为rev标记swap的时候对应的值也要交换).维护的时候d ...
- HDU 5052 Yaoge’s maximum profit 光秃秃的树链拆分 2014 ACM/ICPC Asia Regional Shanghai Online
意甲冠军: 特定n小点的树权. 以下n每一行给出了正确的一点点来表达一个销售点每只鸡价格的格 以下n-1行给出了树的侧 以下Q操作 Q行 u, v, val 从u走v,程中能够买一个鸡腿,然后到后面卖 ...
- codeforces1107G Vasya and Maximum Profit 【模拟】
题目分析: 前缀和啥的模拟一下就行了. 代码: #include<bits/stdc++.h> using namespace std; ; int n,x,d[maxn],sta[max ...
- Codeforces 1107G Vasya and Maximum Profit [单调栈]
洛谷 Codeforces 我竟然能在有生之年踩标算. 思路 首先考虑暴力:枚举左右端点直接计算. 考虑记录\(sum_x=\sum_{i=1}^x c_i\),设选\([l,r]\)时那个奇怪东西的 ...
随机推荐
- HTML5播放器 MediaElement.js 使用方法
目前已经有很多html5播放器可以使用,使用html5播放器可以轻松的在页面中插入媒体视频,从而使我们的web页面变得更加丰富多彩,所以今 天向大家推荐一款非常优秀的html5播放器MediaElem ...
- CSS镂空图片处理
来源:http://www.zhangxinxu.com/wordpress/?p=5267,分享收藏 使用镂空图片,通过CSS改变颜色,达到图片切换的效果,可以同过背景图,然后改变背景色,从而达到图 ...
- shiro 没有权限异常处理
<bean class="org.springframework.web.servlet.handler.SimpleMappingExceptionResolver"> ...
- canal 监控数据库表 快速使用
https://github.com/alibaba/canal 快速开始 https://github.com/alibaba/canal/wiki/QuickStart 注意 1. vim con ...
- Quartz+TopShelf实现Windows服务作业调度
Quartz:首先我贴出来了两段代码(下方),可以看出,首先会根据配置文件(quartz.config),包装出一个Quartz.Core.QuartzScheduler instance,这是一个调 ...
- (4)shiro多个realm
shiro支持多个realm,当设置多个realm的时候,shiro的认证和授权的步骤是怎样的呢. 多个realm认证原理: 发现需要在执行认证的时候,需要策略来处理多个realm存在的情况.默认实现 ...
- session会话管理,与过滤器使用,访问控制
1 用户登录,是否注册用户,在登录处理页面进行用户验证,创建session保存用户名和密码 2否,进入用户注册页面 3是,系统保存该用户的登录信息 4进入要访问的页面 5用户直接访问某个页面, 6系统 ...
- tensor flow 视频
http://v.youku.com/v_show/id_XMTYxMjQ2NTYyNA==.html?spm=a2h1n.8251843.playList.5!19~5~A.siMjNW&f ...
- PCA和SVD(转)
最近突然看到一个问题,PCA和SVD有什么关系?隐约记得自己照猫画虎实现的时候PCA的时候明明用到了SVD啊,但SVD(奇异值分解)和PCA的(特征值分解)貌似差得相当远,由此钻下去搜集了一些资料,把 ...
- Ubuntu 安装 Zabbix 3.2详细步骤
创建 zabbix 用户 因为zabbix 程序的守护进程需要非特权用户,所以需要创建一个 zabbix 用户,即使从 root 用户启动 zabbix 程序,也会自动切换到 zabbix 用户,所以 ...