- #STOCK QUOTES IN NUMBERS FOR MAC SOFTWARE#
- #STOCK QUOTES IN NUMBERS FOR MAC CODE#
- #STOCK QUOTES IN NUMBERS FOR MAC WINDOWS#
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#STOCK QUOTES IN NUMBERS FOR MAC WINDOWS#
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#STOCK QUOTES IN NUMBERS FOR MAC SOFTWARE#
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Stock at multiple Locations With store management software you can keep stock at multiple locations. The software can send you auto alerts to get the material back. We like to check what the smart money thinks first before doing extensive. We also need to look at previous best local min (C# implementation).Keeping this in mind, let's take a look at whether The Macerich Company (NYSE:MAC) is a good investment right now. You can verify that the profit calculation is correct with a simple loop (for simplicity imagine it's within the above function) stock=0Īnother O(n) solution for this task can be done by using local minimum and maximum finding the best deference (profit) between max and min knowing that max should have greater index then min. If ai=m then the profit from stocks bought at the the step is 0: we had decreasing or stable price after that point and did not buy. Note that m is the highest stock price we have seen (from the end).
Should be readable for a C person) def calcprofit(stockvalues):ĭobuy=*len(stockvalues) # 1 for buy, 0 for sellįor i in reversed(range(len(stockvalues))):Įxamples: calcprofit() gives (3, )Ĭalcprofit() gives (197, )Ĭalcprofit() gives
#STOCK QUOTES IN NUMBERS FOR MAC CODE#
Here's the code in C-like python: (I avoided most pythonic stuff. The whole problem is solved with one single reverse loop: calculating both the decisions and the profit of the trade. continue the same way until the beginning. Is it the highest price so far (from all we looked at yet)? - Then sell all, you will not find a better day. Then go to the next day (remember, backwards in time). Is this the highest price so far (from the end), then sell! The last day (where we start reading) you will always sell. Reading from the end, look at price of that day. If you think code is easier to read than words, just skip my explanation, but here goes: Stock trade is easy if your travel backwards in time! To find the sell/buy days you just need to look at each day once: I agree with the logic of your method but there is no need to do recursive processing or global maxima searches. this solution passed 10 of the 11 cases but exceeded the time limit on a last test case (i.e the largest input)Ĭan anyone think of a more efficient solution to this problem? Is there a dynamic programming solution ? The complexity for this turns out to be O(n^2). so we buy stock on day 1 and sell it on day 2 ( profit = 3 ) then we recurse on the remaining days : 1 2 3ī) Max price is 3 ( on day 5) so we keep buying stock on day 3 and day 4 and sell on day 5 ( profit = ( 3*2 - 3 = 3 ) Sell all the stocks on that day and split the array after that day and recurse on the remaining elementsĪ) highest stock price on day 2. Keep buying 1 unit of stock till that day.ī) If that day is the last day then quit:
total profit = 3Ī) Find the day when the stock price was largest. What is the maximum profit you can obtain by planning your trading strategy optimally?**Įxamples ( The input i.e the no of days can vary )ĥ 3 2 => profit = 0 // since the price decreases each day ,the max profit we can make = 0ġ 3 1 2 =>profit = 3 // we buy at 1 sell at 3, then we buy at 1 and sell at 2. Each day, you can either buy one unit of stock, sell any number of stock units you have already bought, or do nothing. You are given the stock prices for a set of days. I was asked this question while interviewing for a startup and saw this again in the recent contest at