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# HOW TO USE 2 STACKS

Here is the problem I am trying to solve.

When a share of common stock of some company is sold, the capital gain (or, sometimes, loss) is the difference between the share's selling price and the price originally paid to buy it. This rule is easy to understand for a single share, but if we sell multiple shares of stock bought over a long period of time, then we must identify the shares actually being sold. A standard accounting principle for identifying which shares of a stock were sold in such a case is to use a FIFO protocol-the shares sold are the ones that have been held the longest (indeed, this is the default method built into several personal finance software packages). For example, suppose we buy 100 shares at \$20 each on day 1, 20 shares at \$24 on day 2, 200 shares at \$36 on day 3, and then sell 150 shares on day 4 at \$30 each. Then applying the FIFO protocol means that of the 150 shares sold, 100 were bought on day 1, 20 were bought on day 2, and 30 were bought on day 3. The capital gain in this case would therefore be 100.10+20.6+30.(&#8722;6), or \$940.

Write a program that takes as input a sequence of transactions of the form

* buy x share(s) at \$y each

or

* sell x share(s) at \$y each,

assuming that the transactions occur on consecutive days and the values x and y are integers. Given this input sequence, the output should be the total capital gain (or loss) for the entire sequence, using the FIFO protocol to identify shares.

My question is how would I define two stacks(or queues)  to match sell orders with buy orders?

Thanks.
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kaushalshah
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1 Solution

Commented:
Sounds like homework?

But to give you an idea:

You have to use one queue. At one end feed in the shares you've bought. The objects you feed in should keep track of share count and prices.
If you want to sell get objects from the other end of queue until you have enough shares from the selling price and the price stored in the object you can calc your gain/loss.

Because normally not all shares from the last object will be used use a queue which allows to peek (instead of remove it) to the last object and subtract as much shares you need from this object.

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Author Commented:
Venabili, I am not asking for code, just ideas.

K
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Author Commented:
JugglerW, I did not understand your last comment.

K
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Commented:
I meant:

If you want to sell 50 shares but the last object in your queue represents 100 shares you can't remove the whole object from the queue but only have to subtract the 50 you need and leave the object in the queue. So you need an operation which allows you to inspect / manipulate the last object without removing it from the queue.

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Commented:
I Need Code For this Assignment Please
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