getting profit in array without using loops

Question

I have something like this in tradeTest.cpp

int main() {
    vector<int> prices{38, 28, 30, 38, 34};

    int profit = bestProfit(prices.begin(), prices.end());

    if (profit == 10) {
        cout << "Profit of 10 is correct\n";
    } else {
        cout << "Profit of " << profit << " is incorrect\n";
    }
}

and currently this in trade.h

template <class Iterator>
int Profit (Iterator begin, Iterator end)
{

}

What I want to do is buy low and sell high without the possibility of going back.

Thanks for your itme.

If they were sorted. Then the low price would be `*begin` and the high price would be `*(end - 1)` — Martin York, Dec 20 '17 at 02:28
@Carcigenicate sorry, I edited the question what I meant is sell high. Thanks — justewe3, Dec 20 '17 at 02:29
You can go back. There’s nothing stopping you from copying the begin iterator and iterating multiple times. — Cris Luengo, Dec 20 '17 at 02:30
Think about it. If you just iterate sequentially one time, how are you going to know to buy at 18 if you don't know if any subsequent values are lower, or sell at 26 if you don't know if any subsequent values are higher? Obviously, you are going to have to sort the input first, or iterate multiple times — Remy Lebeau, Dec 20 '17 at 02:30
@JakeFreeman yes, I think I need to use those two but not sure how — justewe3, Dec 20 '17 at 02:31
Also, you have a random access iterator, you can go backwards with those too. You can access all elements of the array withour limitation. — Cris Luengo, Dec 20 '17 at 02:33
justewe why not just plug it right into the sort method which is part of algo header — Jake Freeman, Dec 20 '17 at 02:34
@CrisLuengo: sure, logically, in this example, the iterators are random-access. But the function doesn't know that, unless that is stipulated as an input requirement, or even forced using SFINAE via `std::enable_if` and `std::iterator_traits::iterator_category`, etc. — Remy Lebeau, Dec 20 '17 at 02:35
@CrisLuengo It was wrong, sorry. I just added it in the question — justewe3, Dec 20 '17 at 02:38
Still you can copy the iterator and iterate multiple times. I would probably loop over all elements, and for each loop over the remaining elements. If you buy at that time, what is the max profit you can get? Then accumulate the max over all start times. O(n^2), but who cares? — Cris Luengo, Dec 20 '17 at 02:40
@CrisLuengo can u post something I begin with if u don't want to post the answer. I am still now sure how copying the iterator will help. — justewe3, Dec 20 '17 at 02:54
I’m typing on a phone, but I could try to write a quick answer... — Cris Luengo, Dec 20 '17 at 02:56

Jake Freeman · Accepted Answer · 2017-12-20T03:37:41.447

2

An easy way to do what you want would be with adding an int max = 0 parameter:

template <class Iterator>
int bestProfit (Iterator begin, Iterator end, int max = 0)
{
    if(begin==end)
    {
        return max;
    }else
    {
        int p = *std::max_element(begin+1, end) - *begin;
        max = std::max(max, p);        
        return bestProfit(begin+1, end, max);
    }

}

int main() {
    vector<int> prices{28, 18, 20, 26, 24};

    int profit = bestProfit(prices.begin(), prices.end());

    if (profit == 8) {
        cout << "Profit of 8 is correct\n";
    } else {
        cout << "Profit of " << profit << " is incorrect\n";
    }
}

edited Dec 20 '17 at 03:37

answered Dec 20 '17 at 02:48

Jake Freeman

1,700
1
8
15

As I said on the other answer, sorting doesnt solve the problem. – Cris Luengo Dec 20 '17 at 02:49
This is wrong the profit is 10 while it should be 26-18 = 8 – justewe3 Dec 20 '17 at 02:51
@CrisLuengo this sorted allows you to get the highest and lowest value. Sorting is the solution is correct – Jake Freeman Dec 20 '17 at 02:51
@justewe3 28 is the highest and 18 is the lowest perhaps you should tell us what 26 is – Jake Freeman Dec 20 '17 at 02:52
@JakeFreeman the thing is that u can't go back on time. I mentioned it in the question. So once u buy u must sell in the future ( the numbers after the number that u buy ( – justewe3 Dec 20 '17 at 02:52
If you sort the values, you lose temporal relationship. If the highest values is the first and the lowest value is the last, their difference is not how much profit you can make. – Cris Luengo Dec 20 '17 at 02:53
@justewe3 updated my answer to efficiently do your task without sort or for loops – Jake Freeman Dec 20 '17 at 03:05
That’s quite nice and compact. +1 – Cris Luengo Dec 20 '17 at 03:08
Actually, come to think of it, this buys lowest, but not necessarily at the point of max return. What if the lowest point is the last? You make 0 profit! – Cris Luengo Dec 20 '17 at 03:09
@CrisLuengo Then the max is the min because it is the elements in the range of the index of the min and the end. The OP requires buying at the lowest point. Just confirmed it returns 0. – Jake Freeman Dec 20 '17 at 03:10
Say the input is 1,2,3,4,5,0. Your answer is 0, but max profit is 4. – Cris Luengo Dec 20 '17 at 03:12
@CrisLuengo is right, if the lowest is in the end. Then it should buy at 20 and sell at 26 – justewe3 Dec 20 '17 at 03:12
@JakeFreeman although this is not working all the times but it is the best form the posted ones. Thanks for the answer – justewe3 Dec 20 '17 at 03:24
@justewe3 I am working on a recursive version which will work in all cases. – Jake Freeman Dec 20 '17 at 03:26
@JakeFreeman I am allowed to only use one function. I am not sure if that will be possible with only one function ( bestProfit ) – justewe3 Dec 20 '17 at 03:27
@justewe3 could I modify what parameters it takes? – Jake Freeman Dec 20 '17 at 03:29
@justewe3 I made it into a recursive function using crisLuengo function and modified it – Jake Freeman Dec 20 '17 at 03:38
@CrisLuengo thank you I used you function to make my recursive function. – Jake Freeman Dec 20 '17 at 03:38

Cris Luengo · Answer 2 · 2017-12-20T04:56:56.133

~~As I suggested in the comments, you might want to do something like this. Sorry I can’t test right now, no guarantees!~~

template <class Iterator>
int bestProfit (Iterator begin, Iterator end)
{
  int profit = 0;
  for(;begin!=end; ++begin){
    int p = *std::max_element(begin+1, end) - *begin;
    profit = std:max(profit, p);
  }
  return profit;
}

Could also be quite elegant with a recursive function, as suggested somewhere in a comment.

The code above should probably skip the last loop iteration, come to think of it. What does std::max_element return if both input iterators are identical?

EDIT:

You already have a working and accepted answer, but just for completeness sake (I guess I'm stubborn like that) here's my version of the recursive function. This does not have any explicit loops, but there is a loop inside of std::max_element, and recursive functions a way of writing a loop. There is no way to visit all elements in a vector or list without a loop!

#include <vector>
#include <algorithm>
#include <iostream>

// Look mama! No `for`!
template <class Iterator>
int bestProfit (Iterator begin, Iterator end)
{
  int buy = *begin;
  ++begin;
  if (begin == end) return 0; // no profit to be had
  int sell = *std::max_element(begin, end);
  int profit = sell - buy; // max profit if we buy now
  return std::max(profit, bestProfit(begin, end)); // will we do better if we wait?
}

int main() {
  std::vector<int> prices1{38, 28, 30, 38, 34};
  if (bestProfit(prices1.begin(), prices1.end()) != 10)
    std::cout << "bad 1!\n";

  std::vector<int> prices2{1, 2, 3, 4, 5, 0};
  if (bestProfit(prices2.begin(), prices2.end()) != 4)
    std::cout << "bad 2!\n";

  std::vector<int> prices3{2, 3, 1, 4, 5, 0};
  if (bestProfit(prices3.begin(), prices3.end()) != 4)
    std::cout << "bad 3!\n";

  std::vector<int> prices4{0, 1, 2, 5, 4, 3};
  if (bestProfit(prices4.begin(), prices4.end()) != 5)
    std::cout << "bad 4!\n";

  std::vector<int> prices5{100, 200, 0, 1, 3};
  if (bestProfit(prices5.begin(), prices5.end()) != 100)
    std::cout << "bad 5!\n";

  std::vector<int> prices6{100, 200, 0, 101, 30};
  if (bestProfit(prices6.begin(), prices6.end()) != 101)
    std::cout << "bad 6!\n";
}

I tried it out but I am getting this error `error: cannot convert ‘__gnu_cxx::__normal_iterator >’ to ‘int’ in initialization ` and it is pointing at `*begin;` Any idea of how to fix it ? — justewe3, Dec 20 '17 at 03:14
Ah, makes sense, sorry, I’ll fix it. It needs a dereference. — Cris Luengo, Dec 20 '17 at 03:15
I think this dereferneces the end iterator in the last loop iteration. I need a better screen to do this right :/ — Cris Luengo, Dec 20 '17 at 03:18
I've just realised that you are using a for loop, the question is saying **without using imperative loop** — justewe3, Dec 20 '17 at 03:19
Thanks for the answer though, it works great but it is using a for loop — justewe3, Dec 20 '17 at 03:19
I guess you need to recode it as a recursive function then. That’s the sneaky way of looping... — Cris Luengo, Dec 20 '17 at 03:20
The problem is that I need to use only one function, so recursion is much of help — justewe3, Dec 20 '17 at 03:21

score 0 · Answer 3 · answered Dec 20 '17 at 02:53

0

It's a simple divine and Conquer algorithm. And also you haven't described the problem correctly i think, sorting the array is wrong. Cause 18-28 will be the maximum but you can't buy at 18 and sell at 28 if the array is at that order

answered Dec 20 '17 at 02:53

Also really before asking stack overflow open your algorithm's book. This is a classic problem for cs universities in algorithms – Dec 20 '17 at 02:54
get a life bro. – justewe3 Dec 20 '17 at 03:23

getting profit in array without using loops

3 Answers3

EDIT: