# Scala algorithm: Find maximum potential profit from an array of stock price

## Algorithm goal

Find the maximum potential profit given an array of stock prices, based on a buy followed by a sell. Where not possible to profit, return None.- Here we look for an efficient solution ($$O(n)$$). This problem is known as:

• On Codility: MaxProfit - Given a log of stock prices compute the maximum possible earning. (100% pass)

## Algorithm in Scala

20 lines of Scala (compatible versions 2.13 & 3.0), showing how concise Scala can be!

def findMaximumProfit(stockPrices: Array[Int]): Option[Int] = {
val maximumSellPricesFromIonward = stockPrices.view
.scanRight(0)({ case (maximumPriceSoFar, dayPrice) =>
Math.max(maximumPriceSoFar, dayPrice)
})
.toArray
val maximumSellPricesAfterI = maximumSellPricesFromIonward.drop(1)
if (stockPrices.length < 2) None
else
stockPrices
.zip(maximumSellPricesAfterI)
})
.max
}

def getPotentialProfit(buyPrice: Int, sellPrice: Int): Option[Int] = {
}


## Test cases in Scala

assert(findMaximumProfit(Array(163, 112, 105, 100, 151)).contains(51))
assert(findMaximumProfit(Array(1)).isEmpty)
assert(findMaximumProfit(Array(1, 2)).contains(1))
assert(findMaximumProfit(Array(2, 1)).isEmpty)


## Explanation

We will derive the solution mathematically, so that we are certain that we are right (I will not even name the algorithm in this explanation because the mathematics is quite important so you can derive the solution for variations of this problem - knowing the algorithm is mostly not enough).

Our final goal is to find out $$\_{\max} = \max\{\_1, \_2, ..., \_{n-1}\}$$, where $$\_d$$ is the maximum profit that that can be gained by can be achieved at buying a stock on day $$d$$ at price $$p_d$$ and selling it any day after $$d$$. $$n-1$$ is because cannot buy-and-sell on the same day. (this is © from www.scala-algorithms.com)

The price at which we sell stock after day $$d$$ is simply the maximum price after day $$d$$ (as to maximise the profit). If we name that $$S_d$$ then it will be $$S_d = \max\{p_{d+1}, p_{d+2}, ..., p_{n}\}$$. Then, $$\_d = S_d - p_d$$.

Notice that $$\max\{a, b, c\} = \max\{\max\{a, b\}, c\}$$ (this is very important in this sort of problems), meaning that $$S_d = \max\{p_{d+1},S_{d+1}\}$$.

Computing $$\_{max} = \max\{S_1-p_1,S_2-p_2, ..., S_{n-1} - p_{n-1}\}$$ directly is $$O(n^2)$$ but because of our earlier relation, we can pre-compute the value of $$S_d$$ from the value of $$S_{d+1}$$ - it just means we have compute from right to left - in fact we can compute the whole array of $$S$$ like that.

After computing $$S_d$$, we already have $$p_d$$ from the original price array, and then we can compute $$\_d$$ from all pairings of $$S_d$$ and $$p_d$$ to eventually find $$\_{max}$$.

In Scala, however, we needn't iterate and mutate - we can utilise functional programming and Scala's powerful collections. Please see the 'Scala concepts' below.

We plug in the relation for $$S_d$$ into a scanRight, then we zip the prices $$p_d$$ with $$S_d$$ to find maximum potential profit at day $$d$$, and then we find the maximum value across all values of $$d$$.

### How can you do .max on a Array[Option[Int]]?!!!

Scala is powerful. It has something called an Ordering which is automatically generated for eg Option so long as there is an Ordering for an Int.

## Scala concepts & Hints

1. ### Drop, Take, dropRight, takeRight

Scala's drop and take methods typically remove or select n items from a collection.

assert(List(1, 2, 3).drop(2) == List(3))

assert(List(1, 2, 3).take(2) == List(1, 2))

assert(List(1, 2, 3).dropRight(2) == List(1))

assert(List(1, 2, 3).takeRight(2) == List(2, 3))

assert((1 to 5).take(2) == (1 to 2))

2. ### Option Type

The 'Option' type is used to describe a computation that either has a result or does not. In Scala, you can 'chain' Option processing, combine with lists and other data structures. For example, you can also turn a pattern-match into a function that return an Option, and vice-versa!

assert(Option(1).flatMap(x => Option(x + 2)) == Option(3))

assert(Option(1).flatMap(x => None) == None)

3. ### Pattern Matching

Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.

assert("Hello World".collect {
case character if Character.isUpperCase(character) => character.toLower
} == "hw")

4. ### scanLeft and scanRight

Scala's scan functions enable you to do folds like foldLeft and foldRight, while collecting the intermediate results

assert(List(1, 2, 3, 4, 5).scanLeft(0)(_ + _) == List(0, 1, 3, 6, 10, 15))

5. ### Stack Safety

Stack safety is present where a function cannot crash due to overflowing the limit of number of recursive calls.

This function will work for n = 5, but will not work for n = 2000 (crash with java.lang.StackOverflowError) - however there is a way to fix it :-)

In Scala Algorithms, we try to write the algorithms in a stack-safe way, where possible, so that when you use the algorithms, they will not crash on large inputs. However, stack-safe implementations are often more complex, and in some cases, overly complex, for the task at hand.

def sum(from: Int, until: Int): Int =
if (from == until) until else from + sum(from + 1, until)

def thisWillSucceed: Int = sum(1, 5)

def thisWillFail: Int = sum(1, 300)

6. ### View

The .view syntax creates a structure that mirrors another structure, until "forced" by an eager operation like .toList, .foreach, .forall, .count.

7. ### Zip

'zip' allows you to combine two lists pair-wise (meaning turn a pair of lists, into a list of pairs)

It can be used over Arrays, Lists, Views, Iterators and other collections.

assert(List(1, 2, 3).zip(List(5, 6, 7)) == List(1 -> 5, 2 -> 6, 3 -> 7))

assert(List(1, 2).zip(List(5, 6, 7)) == List(1 -> 5, 2 -> 6))

assert(List(5, 6).zipWithIndex == List(5 -> 0, 6 -> 1))

assert(List(5, 6).zipAll(List('A'), 9, 'Z') == List(5 -> 'A', 6 -> 'Z'))

assert(List(5).zipAll(List('A', 'B'), 1, 'Z') == List(5 -> 'A', 1 -> 'B'))


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