Greedy-AppleStocks.swift

/*
Writing programming interview questions hasn't made me rich yet ... so I might give up and start trading Apple stocks all day instead.

First, I wanna know how much money I could have made yesterday if I'd been trading Apple stocks all day.

So I grabbed Apple's stock prices from yesterday and put them in an array called stockPrices, where:

The indices are the time (in minutes) past trade opening time, which was 9:30am local time.
The values are the price (in US dollars) of one share of Apple stock at that time.
So if the stock cost $500 at 10:30am, that means stockPrices[60] = 500.

Write an efficient function that takes stockPrices and returns the best profit I could have made from one purchase and one sale of one share of Apple stock yesterday.

For example:
let stockPrices = [10, 7, 5, 8, 11, 9]

getMaxProfit(from: stockPrices)
// returns 6 (buying for $5 and selling for $11)
*/

//Solution does not take in account some of the details mentioned by cake interview but gets close
func getMaxProfit(from stockPricesYesterday: [Int]) -> Int? {
	
	guard stockPrices.count >= 2 else {return nil}
	
	var minPrice:Int?
	var maxPrice = 0
	
	for num in stockPricesYesterday{
		if minPrice == nil {
			minPrice = num
		} else {
			if num < minPrice!{
				minPrice = num
			}
		}
		if num > maxPrice{
			maxPrice = num
		}
	}
	
	return maxPrice - minPrice!
}
let stockPrices = [10, 7, 5, 8, 11, 9]
print(getMaxProfit(from: stockPrices))


//Cake Interview Solution
unc getMaxProfit(from stockPricesYesterday: [Int]) -> Int? {

	guard stockPricesYesterday.count >= 2 else {
		return nil
	}

	// we'll greedily update minPrice and maxProfit, so we initialize
	// them to the first price and the first possible profit
	var minPrice  = stockPricesYesterday[0]
	var maxProfit = stockPricesYesterday[1] - stockPricesYesterday[0]

	// start at the second (index 1) time
	// we can't sell at the first time, since we must buy first,
	// and we can't buy and sell at the same time!
	// if we started at index 0, we'd try to buy *and* sell at time 0.
	// this would give a profit of 0, which is a problem if our
	// maxProfit is supposed to be *negative*--we'd return 0.
	for currentPrice in stockPricesYesterday[1...] {

		// see what our profit would be if we bought at the
		// min price and sold at the current price
		let potentialProfit = currentPrice - minPrice

		// update maxProfit if we can do better
		maxProfit = max(maxProfit, potentialProfit)

		// update minPrice so it's always
		// the lowest price we've seen so far
		minPrice = min(minPrice, currentPrice)
	}

	return maxProfit
}